Example 1: Using the factorization method solve the quadratic equation: x2 +24x +143= 0

Let's rewrite this equation as: x2 +24x +143= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +143x2 while the sum remains +24x
x2 + 13x  + 11x  +143= 0
=> x(x + 13) + 11 (x + 13) = 0
=> (x + 13) (x + 11) = 0
=>  (x + 13) =0  OR (x + 11) = 0
=> x = - 13  OR x = - 11
=> So, by factorization, we solve and find that the roots of the given equation are  - 13 , - 11

Example 2: Using the factorization method solve the quadratic equation: 2x2 -x -231= 0

Let's rewrite this equation as: 2x2 -x -231= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is -462x2 (product of the first and the last term) while the sum remains -x
2x2 + 21x  - 22x  -231= 0
=> x(2x + 21) - 11 (2x + 21) = 0
=> (2x + 21) (x - 22) = 0
=>  (2x + 21) =0  OR (x - 11) = 0
=> x = - 21/2  OR x = + 11
=> So, by factorization, we solve and find that the roots of the given equation are  - 21/2 , + 11

Example 3: Using the factorization method solve the quadratic equation: 15x2 -161x +300= 0

Let's rewrite this equation as: 15x2 -161x +300= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +300x2 while the sum remains -161x
15 x2 - 36x  - 125x  +300= 0
=> 3x(5x - 12) - 25 (5x - 12) = 0
=> (5x - 12) (3x - 25) = 0
=>  (5x - 12) =0  OR (3x - 25) = 0
=> x = + 12/5  OR x = + 25/3
=> So, by factorization, we solve and find that the roots of the given equation are  + 12/5 , + 25/3

Example 4: Using the factorization method solve the quadratic equation: x2 +10x -119= 0

Let's rewrite this equation as: x2 +10x -119= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is -119x2 while the sum remains +10x
x2 + 17x  - 7x  -119= 0
=> x(x + 17) - 7 (x + 17) = 0
=> (x + 17) (x - 7) = 0
=>  (x + 17) =0  OR (x - 7) = 0
=> x = - 17  OR x = + 7
=> So, by factorization, we solve and find that the roots of the given equation are  - 17 , + 7

Example 5: Using the factorization method solve the quadratic equation: x2 -9x -136= 0

Let's rewrite this equation as: x2 -9x -136= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is -136x2 while the sum remains -9x
x2 - 17x  + 8x  -136= 0
=> x(x - 17) + 8 (x - 17) = 0
=> (x - 17) (x + 8) = 0
=>  (x - 17) =0  OR (x + 8) = 0
=> x = + 17  OR x = - 8
=> So, by factorization, we solve and find that the roots of the given equation are  + 17 , - 8

Example 6: Using the factorization method solve the quadratic equation: x2 +3x +2= 0

Let's rewrite this equation as: x2 +3x +2= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +2x2 while the sum remains +3x
x2 + 1x  + 2x  +2= 0
=> x(x + 1) + 2 (x + 1) = 0
=> (x + 1) (x + 2) = 0
=>  (x + 1) =0  OR (x + 2) = 0
=> x = - 1  OR x = - 2
=> So, by factorization, we solve and find that the roots of the given equation are  - 1 , - 2

Example 7: Using the factorization method solve the quadratic equation: x2 -10x -39= 0

Let's rewrite this equation as: x2 -10x -39= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is -39x2 while the sum remains -10x
x2 - 13x  + 3x  -39= 0
=> x(x - 13) + 3 (x - 13) = 0
=> (x - 13) (x + 3) = 0
=>  (x - 13) =0  OR (x + 3) = 0
=> x = + 13  OR x = - 3
=> So, by factorization, we solve and find that the roots of the given equation are  + 13 , - 3

Example 8: Using the factorization method solve the quadratic equation: x2 +23x +102= 0

Let's rewrite this equation as: x2 +23x +102= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +102x2 while the sum remains +23x
x2 + 6x  + 17x  +102= 0
=> x(x + 6) + 17 (x + 6) = 0
=> (x + 6) (x + 17) = 0
=>  (x + 6) =0  OR (x + 17) = 0
=> x = - 6  OR x = - 17
=> So, by factorization, we solve and find that the roots of the given equation are  - 6 , - 17

Example 9: Using the factorization method solve the quadratic equation: x2 -16x +39= 0

Let's rewrite this equation as: x2 -16x +39= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +39x2 while the sum remains -16x
x2 - 3x  - 13x  +39= 0
=> x(x - 3) - 13 (x - 3) = 0
=> (x - 3) (x - 13) = 0
=>  (x - 3) =0  OR (x - 13) = 0
=> x = + 3  OR x = + 13
=> So, by factorization, we solve and find that the roots of the given equation are  + 3 , + 13

Example 10: Using the factorization method solve the quadratic equation: x2 -19x +70= 0

Let's rewrite this equation as: x2 -19x +70= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +70x2 while the sum remains -19x
x2 - 14x  - 5x  +70= 0
=> x(x - 14) - 5 (x - 14) = 0
=> (x - 14) (x - 5) = 0
=>  (x - 14) =0  OR (x - 5) = 0
=> x = + 14  OR x = + 5
=> So, by factorization, we solve and find that the roots of the given equation are  + 14 , + 5

Example 11: Using the factorization method solve the quadratic equation: x2 -18x +17= 0

Let's rewrite this equation as: x2 -18x +17= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +17x2 while the sum remains -18x
x2 - 17x  - 1x  +17= 0
=> x(x - 17) - 1 (x - 17) = 0
=> (x - 17) (x - 1) = 0
=>  (x - 17) =0  OR (x - 1) = 0
=> x = + 17  OR x = + 1
=> So, by factorization, we solve and find that the roots of the given equation are  + 17 , + 1

Example 12: Using the factorization method solve the quadratic equation: x2 +26x +88= 0

Let's rewrite this equation as: x2 +26x +88= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +88x2 while the sum remains +26x
x2 + 4x  + 22x  +88= 0
=> x(x + 4) + 22 (x + 4) = 0
=> (x + 4) (x + 22) = 0
=>  (x + 4) =0  OR (x + 22) = 0
=> x = - 4  OR x = - 22
=> So, by factorization, we solve and find that the roots of the given equation are  - 4 , - 22

Example 13: Using the factorization method solve the quadratic equation: x2 +16x +15= 0

Let's rewrite this equation as: x2 +16x +15= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +15x2 while the sum remains +16x
x2 + 15x  + 1x  +15= 0
=> x(x + 15) + 1 (x + 15) = 0
=> (x + 15) (x + 1) = 0
=>  (x + 15) =0  OR (x + 1) = 0
=> x = - 15  OR x = - 1
=> So, by factorization, we solve and find that the roots of the given equation are  - 15 , - 1

Example 14: Using the factorization method solve the quadratic equation: x2 +25x +46= 0

Let's rewrite this equation as: x2 +25x +46= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +46x2 while the sum remains +25x
x2 + 23x  + 2x  +46= 0
=> x(x + 23) + 2 (x + 23) = 0
=> (x + 23) (x + 2) = 0
=>  (x + 23) =0  OR (x + 2) = 0
=> x = - 23  OR x = - 2
=> So, by factorization, we solve and find that the roots of the given equation are  - 23 , - 2

Example 15: Using the factorization method solve the quadratic equation: x2 -17x +72= 0

Let's rewrite this equation as: x2 -17x +72= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +72x2 while the sum remains -17x
x2 - 9x  - 8x  +72= 0
=> x(x - 9) - 8 (x - 9) = 0
=> (x - 9) (x - 8) = 0
=>  (x - 9) =0  OR (x - 8) = 0
=> x = + 9  OR x = + 8
=> So, by factorization, we solve and find that the roots of the given equation are  + 9 , + 8

Example 16: Using the factorization method solve the quadratic equation: x2 -4x -117= 0

Let's rewrite this equation as: x2 -4x -117= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is -117x2 while the sum remains -4x
x2 - 13x  + 9x  -117= 0
=> x(x - 13) + 9 (x - 13) = 0
=> (x - 13) (x + 9) = 0
=>  (x - 13) =0  OR (x + 9) = 0
=> x = + 13  OR x = - 9
=> So, by factorization, we solve and find that the roots of the given equation are  + 13 , - 9

Example 17: Using the factorization method solve the quadratic equation: x2 +23x +22= 0

Let's rewrite this equation as: x2 +23x +22= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +22x2 while the sum remains +23x
x2 + 1x  + 22x  +22= 0
=> x(x + 1) + 22 (x + 1) = 0
=> (x + 1) (x + 22) = 0
=>  (x + 1) =0  OR (x + 22) = 0
=> x = - 1  OR x = - 22
=> So, by factorization, we solve and find that the roots of the given equation are  - 1 , - 22

Example 18: Using the factorization method solve the quadratic equation: x2 -10x -56= 0

Let's rewrite this equation as: x2 -10x -56= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is -56x2 while the sum remains -10x
x2 - 14x  + 4x  -56= 0
=> x(x - 14) + 4 (x - 14) = 0
=> (x - 14) (x + 4) = 0
=>  (x - 14) =0  OR (x + 4) = 0
=> x = + 14  OR x = - 4
=> So, by factorization, we solve and find that the roots of the given equation are  + 14 , - 4

Example 19: Using the factorization method solve the quadratic equation: x2 -33x +266= 0

Let's rewrite this equation as: x2 -33x +266= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +266x2 while the sum remains -33x
x2 - 14x  - 19x  +266= 0
=> x(x - 14) - 19 (x - 14) = 0
=> (x - 14) (x - 19) = 0
=>  (x - 14) =0  OR (x - 19) = 0
=> x = + 14  OR x = + 19
=> So, by factorization, we solve and find that the roots of the given equation are  + 14 , + 19

Example 20: Using the factorization method solve the quadratic equation: x2 +26x +133= 0

Let's rewrite this equation as: x2 +26x +133= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +133x2 while the sum remains +26x
x2 + 7x  + 19x  +133= 0
=> x(x + 7) + 19 (x + 7) = 0
=> (x + 7) (x + 19) = 0
=>  (x + 7) =0  OR (x + 19) = 0
=> x = - 7  OR x = - 19
=> So, by factorization, we solve and find that the roots of the given equation are  - 7 , - 19

Example 21: Using the factorization method solve the quadratic equation: x2 +0x -169= 0

Let's rewrite this equation as: x2 +0x -169= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is -169x2 while the sum remains +0x
x2 - 13x  + 13x  -169= 0
=> x(x - 13) + 13 (x - 13) = 0
=> (x - 13) (x + 13) = 0
=>  (x - 13) =0  OR (x + 13) = 0
=> x = + 13  OR x = - 13
=> So, by factorization, we solve and find that the roots of the given equation are  + 13 , - 13

Example 22: Using the factorization method solve the quadratic equation: x2 -11x -350= 0

Let's rewrite this equation as: x2 -11x -350= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is -350x2 while the sum remains -11x
x2 - 25x  + 14x  -350= 0
=> x(x - 25) + 14 (x - 25) = 0
=> (x - 25) (x + 14) = 0
=>  (x - 25) =0  OR (x + 14) = 0
=> x = + 25  OR x = - 14
=> So, by factorization, we solve and find that the roots of the given equation are  + 25 , - 14

Example 23: Using the factorization method solve the quadratic equation: x2 +25x +126= 0

Let's rewrite this equation as: x2 +25x +126= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +126x2 while the sum remains +25x
x2 + 7x  + 18x  +126= 0
=> x(x + 7) + 18 (x + 7) = 0
=> (x + 7) (x + 18) = 0
=>  (x + 7) =0  OR (x + 18) = 0
=> x = - 7  OR x = - 18
=> So, by factorization, we solve and find that the roots of the given equation are  - 7 , - 18

Example 24: Using the factorization method solve the quadratic equation: x2 +12x +11= 0

Let's rewrite this equation as: x2 +12x +11= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +11x2 while the sum remains +12x
x2 + 1x  + 11x  +11= 0
=> x(x + 1) + 11 (x + 1) = 0
=> (x + 1) (x + 11) = 0
=>  (x + 1) =0  OR (x + 11) = 0
=> x = - 1  OR x = - 11
=> So, by factorization, we solve and find that the roots of the given equation are  - 1 , - 11

Example 25: Using the factorization method solve the quadratic equation: x2 +45x +506= 0

Let's rewrite this equation as: x2 +45x +506= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +506x2 while the sum remains +45x
x2 + 22x  + 23x  +506= 0
=> x(x + 22) + 23 (x + 22) = 0
=> (x + 22) (x + 23) = 0
=>  (x + 22) =0  OR (x + 23) = 0
=> x = - 22  OR x = - 23
=> So, by factorization, we solve and find that the roots of the given equation are  - 22 , - 23

Example 26: Using the factorization method solve the quadratic equation: x2 +0x -121= 0

Let's rewrite this equation as: x2 +0x -121= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is -121x2 while the sum remains +0x
x2 - 11x  + 11x  -121= 0
=> x(x - 11) + 11 (x - 11) = 0
=> (x - 11) (x + 11) = 0
=>  (x - 11) =0  OR (x + 11) = 0
=> x = + 11  OR x = - 11
=> So, by factorization, we solve and find that the roots of the given equation are  + 11 , - 11

Example 27: Using the factorization method solve the quadratic equation: x2 +22x +57= 0

Let's rewrite this equation as: x2 +22x +57= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +57x2 while the sum remains +22x
x2 + 19x  + 3x  +57= 0
=> x(x + 19) + 3 (x + 19) = 0
=> (x + 19) (x + 3) = 0
=>  (x + 19) =0  OR (x + 3) = 0
=> x = - 19  OR x = - 3
=> So, by factorization, we solve and find that the roots of the given equation are  - 19 , - 3

Example 28: Using the factorization method solve the quadratic equation: x2 -30x +144= 0

Let's rewrite this equation as: x2 -30x +144= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +144x2 while the sum remains -30x
x2 - 24x  - 6x  +144= 0
=> x(x - 24) - 6 (x - 24) = 0
=> (x - 24) (x - 6) = 0
=>  (x - 24) =0  OR (x - 6) = 0
=> x = + 24  OR x = + 6
=> So, by factorization, we solve and find that the roots of the given equation are  + 24 , + 6

Example 29: Using the factorization method solve the quadratic equation: x2 -26x +48= 0

Let's rewrite this equation as: x2 -26x +48= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +48x2 while the sum remains -26x
x2 - 2x  - 24x  +48= 0
=> x(x - 2) - 24 (x - 2) = 0
=> (x - 2) (x - 24) = 0
=>  (x - 2) =0  OR (x - 24) = 0
=> x = + 2  OR x = + 24
=> So, by factorization, we solve and find that the roots of the given equation are  + 2 , + 24

Example 30: Using the factorization method solve the quadratic equation: x2 -28x +195= 0

Let's rewrite this equation as: x2 -28x +195= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +195x2 while the sum remains -28x
x2 - 15x  - 13x  +195= 0
=> x(x - 15) - 13 (x - 15) = 0
=> (x - 15) (x - 13) = 0
=>  (x - 15) =0  OR (x - 13) = 0
=> x = + 15  OR x = + 13
=> So, by factorization, we solve and find that the roots of the given equation are  + 15 , + 13

Example 31: Using the factorization method solve the quadratic equation: x2 +15x +50= 0

Let's rewrite this equation as: x2 +15x +50= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +50x2 while the sum remains +15x
x2 + 10x  + 5x  +50= 0
=> x(x + 10) + 5 (x + 10) = 0
=> (x + 10) (x + 5) = 0
=>  (x + 10) =0  OR (x + 5) = 0
=> x = - 10  OR x = - 5
=> So, by factorization, we solve and find that the roots of the given equation are  - 10 , - 5

Example 32: Using the factorization method solve the quadratic equation: x2 -5x -374= 0

Let's rewrite this equation as: x2 -5x -374= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is -374x2 while the sum remains -5x
x2 + 17x  - 22x  -374= 0
=> x(x + 17) - 22 (x + 17) = 0
=> (x + 17) (x - 22) = 0
=>  (x + 17) =0  OR (x - 22) = 0
=> x = - 17  OR x = + 22
=> So, by factorization, we solve and find that the roots of the given equation are  - 17 , + 22

Example 33: Using the factorization method solve the quadratic equation: x2 -4x -60= 0

Let's rewrite this equation as: x2 -4x -60= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is -60x2 while the sum remains -4x
x2 + 6x  - 10x  -60= 0
=> x(x + 6) - 10 (x + 6) = 0
=> (x + 6) (x - 10) = 0
=>  (x + 6) =0  OR (x - 10) = 0
=> x = - 6  OR x = + 10
=> So, by factorization, we solve and find that the roots of the given equation are  - 6 , + 10

Example 34: Using the factorization method solve the quadratic equation: x2 +12x +35= 0

Let's rewrite this equation as: x2 +12x +35= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +35x2 while the sum remains +12x
x2 + 5x  + 7x  +35= 0
=> x(x + 5) + 7 (x + 5) = 0
=> (x + 5) (x + 7) = 0
=>  (x + 5) =0  OR (x + 7) = 0
=> x = - 5  OR x = - 7
=> So, by factorization, we solve and find that the roots of the given equation are  - 5 , - 7

Example 35: Using the factorization method solve the quadratic equation: x2 +27x +170= 0

Let's rewrite this equation as: x2 +27x +170= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +170x2 while the sum remains +27x
x2 + 10x  + 17x  +170= 0
=> x(x + 10) + 17 (x + 10) = 0
=> (x + 10) (x + 17) = 0
=>  (x + 10) =0  OR (x + 17) = 0
=> x = - 10  OR x = - 17
=> So, by factorization, we solve and find that the roots of the given equation are  - 10 , - 17

Example 36: Using the factorization method solve the quadratic equation: x2 -9x -360= 0

Let's rewrite this equation as: x2 -9x -360= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is -360x2 while the sum remains -9x
x2 - 24x  + 15x  -360= 0
=> x(x - 24) + 15 (x - 24) = 0
=> (x - 24) (x + 15) = 0
=>  (x - 24) =0  OR (x + 15) = 0
=> x = + 24  OR x = - 15
=> So, by factorization, we solve and find that the roots of the given equation are  + 24 , - 15

Example 37: Using the factorization method solve the quadratic equation: x2 +17x +60= 0

Let's rewrite this equation as: x2 +17x +60= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +60x2 while the sum remains +17x
x2 + 5x  + 12x  +60= 0
=> x(x + 5) + 12 (x + 5) = 0
=> (x + 5) (x + 12) = 0
=>  (x + 5) =0  OR (x + 12) = 0
=> x = - 5  OR x = - 12
=> So, by factorization, we solve and find that the roots of the given equation are  - 5 , - 12

Example 38: Using the factorization method solve the quadratic equation: x2 +13x -198= 0

Let's rewrite this equation as: x2 +13x -198= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is -198x2 while the sum remains +13x
x2 + 22x  - 9x  -198= 0
=> x(x + 22) - 9 (x + 22) = 0
=> (x + 22) (x - 9) = 0
=>  (x + 22) =0  OR (x - 9) = 0
=> x = - 22  OR x = + 9
=> So, by factorization, we solve and find that the roots of the given equation are  - 22 , + 9

Example 39: Using the factorization method solve the quadratic equation: x2 +0x -289= 0

Let's rewrite this equation as: x2 +0x -289= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is -289x2 while the sum remains +0x
x2 - 17x  + 17x  -289= 0
=> x(x - 17) + 17 (x - 17) = 0
=> (x - 17) (x + 17) = 0
=>  (x - 17) =0  OR (x + 17) = 0
=> x = + 17  OR x = - 17
=> So, by factorization, we solve and find that the roots of the given equation are  + 17 , - 17

Example 40: Using the factorization method solve the quadratic equation: x2 +24x +128= 0

Let's rewrite this equation as: x2 +24x +128= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +128x2 while the sum remains +24x
x2 + 8x  + 16x  +128= 0
=> x(x + 8) + 16 (x + 8) = 0
=> (x + 8) (x + 16) = 0
=>  (x + 8) =0  OR (x + 16) = 0
=> x = - 8  OR x = - 16
=> So, by factorization, we solve and find that the roots of the given equation are  - 8 , - 16

Example 41: Using the factorization method solve the quadratic equation: x2 +12x +27= 0

Let's rewrite this equation as: x2 +12x +27= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +27x2 while the sum remains +12x
x2 + 9x  + 3x  +27= 0
=> x(x + 9) + 3 (x + 9) = 0
=> (x + 9) (x + 3) = 0
=>  (x + 9) =0  OR (x + 3) = 0
=> x = - 9  OR x = - 3
=> So, by factorization, we solve and find that the roots of the given equation are  - 9 , - 3

Example 42: Using the factorization method solve the quadratic equation: x2 +18x +77= 0

Let's rewrite this equation as: x2 +18x +77= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +77x2 while the sum remains +18x
x2 + 7x  + 11x  +77= 0
=> x(x + 7) + 11 (x + 7) = 0
=> (x + 7) (x + 11) = 0
=>  (x + 7) =0  OR (x + 11) = 0
=> x = - 7  OR x = - 11
=> So, by factorization, we solve and find that the roots of the given equation are  - 7 , - 11

Example 43: Using the factorization method solve the quadratic equation: x2 -9x -10= 0

Let's rewrite this equation as: x2 -9x -10= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is -10x2 while the sum remains -9x
x2 + 1x  - 10x  -10= 0
=> x(x + 1) - 10 (x + 1) = 0
=> (x + 1) (x - 10) = 0
=>  (x + 1) =0  OR (x - 10) = 0
=> x = - 1  OR x = + 10
=> So, by factorization, we solve and find that the roots of the given equation are  - 1 , + 10

Example 44: Using the factorization method solve the quadratic equation: x2 +43x +456= 0

Let's rewrite this equation as: x2 +43x +456= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +456x2 while the sum remains +43x
x2 + 24x  + 19x  +456= 0
=> x(x + 24) + 19 (x + 24) = 0
=> (x + 24) (x + 19) = 0
=>  (x + 24) =0  OR (x + 19) = 0
=> x = - 24  OR x = - 19
=> So, by factorization, we solve and find that the roots of the given equation are  - 24 , - 19

Example 45: Using the factorization method solve the quadratic equation: x2 +13x +12= 0

Let's rewrite this equation as: x2 +13x +12= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +12x2 while the sum remains +13x
x2 + 12x  + 1x  +12= 0
=> x(x + 12) + 1 (x + 12) = 0
=> (x + 12) (x + 1) = 0
=>  (x + 12) =0  OR (x + 1) = 0
=> x = - 12  OR x = - 1
=> So, by factorization, we solve and find that the roots of the given equation are  - 12 , - 1

Example 46: Using the factorization method solve the quadratic equation: x2 -11x -12= 0

Let's rewrite this equation as: x2 -11x -12= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is -12x2 while the sum remains -11x
x2 - 12x  + 1x  -12= 0
=> x(x - 12) + 1 (x - 12) = 0
=> (x - 12) (x + 1) = 0
=>  (x - 12) =0  OR (x + 1) = 0
=> x = + 12  OR x = - 1
=> So, by factorization, we solve and find that the roots of the given equation are  + 12 , - 1

Example 47: Using the factorization method solve the quadratic equation: x2 +8x -153= 0

Let's rewrite this equation as: x2 +8x -153= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is -153x2 while the sum remains +8x
x2 + 17x  - 9x  -153= 0
=> x(x + 17) - 9 (x + 17) = 0
=> (x + 17) (x - 9) = 0
=>  (x + 17) =0  OR (x - 9) = 0
=> x = - 17  OR x = + 9
=> So, by factorization, we solve and find that the roots of the given equation are  - 17 , + 9

Example 48: Using the factorization method solve the quadratic equation: x2 +26x +144= 0

Let's rewrite this equation as: x2 +26x +144= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +144x2 while the sum remains +26x
x2 + 8x  + 18x  +144= 0
=> x(x + 8) + 18 (x + 8) = 0
=> (x + 8) (x + 18) = 0
=>  (x + 8) =0  OR (x + 18) = 0
=> x = - 8  OR x = - 18
=> So, by factorization, we solve and find that the roots of the given equation are  - 8 , - 18

Example 49: Using the factorization method solve the quadratic equation: x2 +24x +119= 0

Let's rewrite this equation as: x2 +24x +119= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +119x2 while the sum remains +24x
x2 + 7x  + 17x  +119= 0
=> x(x + 7) + 17 (x + 7) = 0
=> (x + 7) (x + 17) = 0
=>  (x + 7) =0  OR (x + 17) = 0
=> x = - 7  OR x = - 17
=> So, by factorization, we solve and find that the roots of the given equation are  - 7 , - 17

Example 50: Using the factorization method solve the quadratic equation: x2 +8x -9= 0

Let's rewrite this equation as: x2 +8x -9= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is -9x2 while the sum remains +8x
x2 - 1x  + 9x  -9= 0
=> x(x - 1) + 9 (x - 1) = 0
=> (x - 1) (x + 9) = 0
=>  (x - 1) =0  OR (x + 9) = 0
=> x = + 1  OR x = - 9
=> So, by factorization, we solve and find that the roots of the given equation are  + 1 , - 9

Example 51: Using the factorization method solve the quadratic equation: x2 +40x +384= 0

Let's rewrite this equation as: x2 +40x +384= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +384x2 while the sum remains +40x
x2 + 24x  + 16x  +384= 0
=> x(x + 24) + 16 (x + 24) = 0
=> (x + 24) (x + 16) = 0
=>  (x + 24) =0  OR (x + 16) = 0
=> x = - 24  OR x = - 16
=> So, by factorization, we solve and find that the roots of the given equation are  - 24 , - 16

Example 52: Using the factorization method solve the quadratic equation: x2 +4x -252= 0

Let's rewrite this equation as: x2 +4x -252= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is -252x2 while the sum remains +4x
x2 - 14x  + 18x  -252= 0
=> x(x - 14) + 18 (x - 14) = 0
=> (x - 14) (x + 18) = 0
=>  (x - 14) =0  OR (x + 18) = 0
=> x = + 14  OR x = - 18
=> So, by factorization, we solve and find that the roots of the given equation are  + 14 , - 18

Example 53: Using the factorization method solve the quadratic equation: x2 +10x +9= 0

Let's rewrite this equation as: x2 +10x +9= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +9x2 while the sum remains +10x
x2 + 9x  + 1x  +9= 0
=> x(x + 9) + 1 (x + 9) = 0
=> (x + 9) (x + 1) = 0
=>  (x + 9) =0  OR (x + 1) = 0
=> x = - 9  OR x = - 1
=> So, by factorization, we solve and find that the roots of the given equation are  - 9 , - 1

Example 54: Using the factorization method solve the quadratic equation: x2 +x -506= 0

Let's rewrite this equation as: x2 +x -506= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is -506x2 while the sum remains +x
x2 - 22x  + 23x  -506= 0
=> x(x - 22) + 23 (x - 22) = 0
=> (x - 22) (x + 23) = 0
=>  (x - 22) =0  OR (x + 23) = 0
=> x = + 22  OR x = - 23
=> So, by factorization, we solve and find that the roots of the given equation are  + 22 , - 23

Example 55: Using the factorization method solve the quadratic equation: x2 +14x +49= 0

Let's rewrite this equation as: x2 +14x +49= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +49x2 while the sum remains +14x
x2 + 7x  + 7x  +49= 0
=> x(x + 7) + 7 (x + 7) = 0
=> (x + 7) (x + 7) = 0
=>  (x + 7) =0  OR (x + 7) = 0
=> x = - 7  OR x = - 7
=> So, by factorization, we solve and find that the roots of the given equation are  - 7 , - 7

Example 56: Using the factorization method solve the quadratic equation: x2 -18x -144= 0

Let's rewrite this equation as: x2 -18x -144= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is -144x2 while the sum remains -18x
x2 - 24x  + 6x  -144= 0
=> x(x - 24) + 6 (x - 24) = 0
=> (x - 24) (x + 6) = 0
=>  (x - 24) =0  OR (x + 6) = 0
=> x = + 24  OR x = - 6
=> So, by factorization, we solve and find that the roots of the given equation are  + 24 , - 6

Example 57: Using the factorization method solve the quadratic equation: x2 -5x -50= 0

Let's rewrite this equation as: x2 -5x -50= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is -50x2 while the sum remains -5x
x2 + 5x  - 10x  -50= 0
=> x(x + 5) - 10 (x + 5) = 0
=> (x + 5) (x - 10) = 0
=>  (x + 5) =0  OR (x - 10) = 0
=> x = - 5  OR x = + 10
=> So, by factorization, we solve and find that the roots of the given equation are  - 5 , + 10

Example 58: Using the factorization method solve the quadratic equation: x2 +5x -84= 0

Let's rewrite this equation as: x2 +5x -84= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is -84x2 while the sum remains +5x
x2 - 7x  + 12x  -84= 0
=> x(x - 7) + 12 (x - 7) = 0
=> (x - 7) (x + 12) = 0
=>  (x - 7) =0  OR (x + 12) = 0
=> x = + 7  OR x = - 12
=> So, by factorization, we solve and find that the roots of the given equation are  + 7 , - 12

Example 59: Using the factorization method solve the quadratic equation: x2 -21x -22= 0

Let's rewrite this equation as: x2 -21x -22= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is -22x2 while the sum remains -21x
x2 - 22x  + 1x  -22= 0
=> x(x - 22) + 1 (x - 22) = 0
=> (x - 22) (x + 1) = 0
=>  (x - 22) =0  OR (x + 1) = 0
=> x = + 22  OR x = - 1
=> So, by factorization, we solve and find that the roots of the given equation are  + 22 , - 1

Example 60: Using the factorization method solve the quadratic equation: x2 +36x +308= 0

Let's rewrite this equation as: x2 +36x +308= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +308x2 while the sum remains +36x
x2 + 22x  + 14x  +308= 0
=> x(x + 22) + 14 (x + 22) = 0
=> (x + 22) (x + 14) = 0
=>  (x + 22) =0  OR (x + 14) = 0
=> x = - 22  OR x = - 14
=> So, by factorization, we solve and find that the roots of the given equation are  - 22 , - 14

Example 61: Using the factorization method solve the quadratic equation: x2 +25x +24= 0

Let's rewrite this equation as: x2 +25x +24= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +24x2 while the sum remains +25x
x2 + 24x  + 1x  +24= 0
=> x(x + 24) + 1 (x + 24) = 0
=> (x + 24) (x + 1) = 0
=>  (x + 24) =0  OR (x + 1) = 0
=> x = - 24  OR x = - 1
=> So, by factorization, we solve and find that the roots of the given equation are  - 24 , - 1

Example 62: Using the factorization method solve the quadratic equation: x2 -4x -221= 0

Let's rewrite this equation as: x2 -4x -221= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is -221x2 while the sum remains -4x
x2 + 13x  - 17x  -221= 0
=> x(x + 13) - 17 (x + 13) = 0
=> (x + 13) (x - 17) = 0
=>  (x + 13) =0  OR (x - 17) = 0
=> x = - 13  OR x = + 17
=> So, by factorization, we solve and find that the roots of the given equation are  - 13 , + 17

Example 63: Using the factorization method solve the quadratic equation: x2 -15x +36= 0

Let's rewrite this equation as: x2 -15x +36= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +36x2 while the sum remains -15x
x2 - 3x  - 12x  +36= 0
=> x(x - 3) - 12 (x - 3) = 0
=> (x - 3) (x - 12) = 0
=>  (x - 3) =0  OR (x - 12) = 0
=> x = + 3  OR x = + 12
=> So, by factorization, we solve and find that the roots of the given equation are  + 3 , + 12

Example 64: Using the factorization method solve the quadratic equation: x2 -7x -408= 0

Let's rewrite this equation as: x2 -7x -408= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is -408x2 while the sum remains -7x
x2 - 24x  + 17x  -408= 0
=> x(x - 24) + 17 (x - 24) = 0
=> (x - 24) (x + 17) = 0
=>  (x - 24) =0  OR (x + 17) = 0
=> x = + 24  OR x = - 17
=> So, by factorization, we solve and find that the roots of the given equation are  + 24 , - 17

Example 65: Using the factorization method solve the quadratic equation: x2 -23x +112= 0

Let's rewrite this equation as: x2 -23x +112= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +112x2 while the sum remains -23x
x2 - 7x  - 16x  +112= 0
=> x(x - 7) - 16 (x - 7) = 0
=> (x - 7) (x - 16) = 0
=>  (x - 7) =0  OR (x - 16) = 0
=> x = + 7  OR x = + 16
=> So, by factorization, we solve and find that the roots of the given equation are  + 7 , + 16

Example 66: Using the factorization method solve the quadratic equation: x2 +11x +24= 0

Let's rewrite this equation as: x2 +11x +24= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +24x2 while the sum remains +11x
x2 + 3x  + 8x  +24= 0
=> x(x + 3) + 8 (x + 3) = 0
=> (x + 3) (x + 8) = 0
=>  (x + 3) =0  OR (x + 8) = 0
=> x = - 3  OR x = - 8
=> So, by factorization, we solve and find that the roots of the given equation are  - 3 , - 8

Example 67: Using the factorization method solve the quadratic equation: x2 +22x +105= 0

Let's rewrite this equation as: x2 +22x +105= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +105x2 while the sum remains +22x
x2 + 15x  + 7x  +105= 0
=> x(x + 15) + 7 (x + 15) = 0
=> (x + 15) (x + 7) = 0
=>  (x + 15) =0  OR (x + 7) = 0
=> x = - 15  OR x = - 7
=> So, by factorization, we solve and find that the roots of the given equation are  - 15 , - 7

Example 68: Using the factorization method solve the quadratic equation: x2 -6x -432= 0

Let's rewrite this equation as: x2 -6x -432= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is -432x2 while the sum remains -6x
x2 + 18x  - 24x  -432= 0
=> x(x + 18) - 24 (x + 18) = 0
=> (x + 18) (x - 24) = 0
=>  (x + 18) =0  OR (x - 24) = 0
=> x = - 18  OR x = + 24
=> So, by factorization, we solve and find that the roots of the given equation are  - 18 , + 24

Example 69: Using the factorization method solve the quadratic equation: x2 +21x +38= 0

Let's rewrite this equation as: x2 +21x +38= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +38x2 while the sum remains +21x
x2 + 19x  + 2x  +38= 0
=> x(x + 19) + 2 (x + 19) = 0
=> (x + 19) (x + 2) = 0
=>  (x + 19) =0  OR (x + 2) = 0
=> x = - 19  OR x = - 2
=> So, by factorization, we solve and find that the roots of the given equation are  - 19 , - 2

Example 70: Using the factorization method solve the quadratic equation: x2 -10x -39= 0

Let's rewrite this equation as: x2 -10x -39= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is -39x2 while the sum remains -10x
x2 + 3x  - 13x  -39= 0
=> x(x + 3) - 13 (x + 3) = 0
=> (x + 3) (x - 13) = 0
=>  (x + 3) =0  OR (x - 13) = 0
=> x = - 3  OR x = + 13
=> So, by factorization, we solve and find that the roots of the given equation are  - 3 , + 13

Example 71: Using the factorization method solve the quadratic equation: x2 +7x +6= 0

Let's rewrite this equation as: x2 +7x +6= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +6x2 while the sum remains +7x
x2 + 1x  + 6x  +6= 0
=> x(x + 1) + 6 (x + 1) = 0
=> (x + 1) (x + 6) = 0
=>  (x + 1) =0  OR (x + 6) = 0
=> x = - 1  OR x = - 6
=> So, by factorization, we solve and find that the roots of the given equation are  - 1 , - 6

Example 72: Using the factorization method solve the quadratic equation: x2 -22x +40= 0

Let's rewrite this equation as: x2 -22x +40= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +40x2 while the sum remains -22x
x2 - 20x  - 2x  +40= 0
=> x(x - 20) - 2 (x - 20) = 0
=> (x - 20) (x - 2) = 0
=>  (x - 20) =0  OR (x - 2) = 0
=> x = + 20  OR x = + 2
=> So, by factorization, we solve and find that the roots of the given equation are  + 20 , + 2

Example 73: Using the factorization method solve the quadratic equation: x2 -10x -299= 0

Let's rewrite this equation as: x2 -10x -299= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is -299x2 while the sum remains -10x
x2 - 23x  + 13x  -299= 0
=> x(x - 23) + 13 (x - 23) = 0
=> (x - 23) (x + 13) = 0
=>  (x - 23) =0  OR (x + 13) = 0
=> x = + 23  OR x = - 13
=> So, by factorization, we solve and find that the roots of the given equation are  + 23 , - 13

Example 74: Using the factorization method solve the quadratic equation: x2 +17x +16= 0

Let's rewrite this equation as: x2 +17x +16= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +16x2 while the sum remains +17x
x2 + 16x  + 1x  +16= 0
=> x(x + 16) + 1 (x + 16) = 0
=> (x + 16) (x + 1) = 0
=>  (x + 16) =0  OR (x + 1) = 0
=> x = - 16  OR x = - 1
=> So, by factorization, we solve and find that the roots of the given equation are  - 16 , - 1

Example 75: Using the factorization method solve the quadratic equation: x2 -2x -99= 0

Let's rewrite this equation as: x2 -2x -99= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is -99x2 while the sum remains -2x
x2 + 9x  - 11x  -99= 0
=> x(x + 9) - 11 (x + 9) = 0
=> (x + 9) (x - 11) = 0
=>  (x + 9) =0  OR (x - 11) = 0
=> x = - 9  OR x = + 11
=> So, by factorization, we solve and find that the roots of the given equation are  - 9 , + 11

Example 76: Using the factorization method solve the quadratic equation: x2 -29x +120= 0

Let's rewrite this equation as: x2 -29x +120= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +120x2 while the sum remains -29x
x2 - 5x  - 24x  +120= 0
=> x(x - 5) - 24 (x - 5) = 0
=> (x - 5) (x - 24) = 0
=>  (x - 5) =0  OR (x - 24) = 0
=> x = + 5  OR x = + 24
=> So, by factorization, we solve and find that the roots of the given equation are  + 5 , + 24

Example 77: Using the factorization method solve the quadratic equation: x2 +14x -147= 0

Let's rewrite this equation as: x2 +14x -147= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is -147x2 while the sum remains +14x
x2 - 7x  + 21x  -147= 0
=> x(x - 7) + 21 (x - 7) = 0
=> (x - 7) (x + 21) = 0
=>  (x - 7) =0  OR (x + 21) = 0
=> x = + 7  OR x = - 21
=> So, by factorization, we solve and find that the roots of the given equation are  + 7 , - 21

Example 78: Using the factorization method solve the quadratic equation: x2 +8x -209= 0

Let's rewrite this equation as: x2 +8x -209= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is -209x2 while the sum remains +8x
x2 - 11x  + 19x  -209= 0
=> x(x - 11) + 19 (x - 11) = 0
=> (x - 11) (x + 19) = 0
=>  (x - 11) =0  OR (x + 19) = 0
=> x = + 11  OR x = - 19
=> So, by factorization, we solve and find that the roots of the given equation are  + 11 , - 19

Example 79: Using the factorization method solve the quadratic equation: x2 +5x -36= 0

Let's rewrite this equation as: x2 +5x -36= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is -36x2 while the sum remains +5x
x2 - 4x  + 9x  -36= 0
=> x(x - 4) + 9 (x - 4) = 0
=> (x - 4) (x + 9) = 0
=>  (x - 4) =0  OR (x + 9) = 0
=> x = + 4  OR x = - 9
=> So, by factorization, we solve and find that the roots of the given equation are  + 4 , - 9

Example 80: Using the factorization method solve the quadratic equation: x2 -14x -240= 0

Let's rewrite this equation as: x2 -14x -240= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is -240x2 while the sum remains -14x
x2 + 10x  - 24x  -240= 0
=> x(x + 10) - 24 (x + 10) = 0
=> (x + 10) (x - 24) = 0
=>  (x + 10) =0  OR (x - 24) = 0
=> x = - 10  OR x = + 24
=> So, by factorization, we solve and find that the roots of the given equation are  - 10 , + 24

Example 81: Using the factorization method solve the quadratic equation: x2 -42x +432= 0

Let's rewrite this equation as: x2 -42x +432= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +432x2 while the sum remains -42x
x2 - 24x  - 18x  +432= 0
=> x(x - 24) - 18 (x - 24) = 0
=> (x - 24) (x - 18) = 0
=>  (x - 24) =0  OR (x - 18) = 0
=> x = + 24  OR x = + 18
=> So, by factorization, we solve and find that the roots of the given equation are  + 24 , + 18

Example 82: Using the factorization method solve the quadratic equation: x2 +0x -169= 0

Let's rewrite this equation as: x2 +0x -169= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is -169x2 while the sum remains +0x
x2 - 13x  + 13x  -169= 0
=> x(x - 13) + 13 (x - 13) = 0
=> (x - 13) (x + 13) = 0
=>  (x - 13) =0  OR (x + 13) = 0
=> x = + 13  OR x = - 13
=> So, by factorization, we solve and find that the roots of the given equation are  + 13 , - 13

Example 83: Using the factorization method solve the quadratic equation: x2 -19x +78= 0

Let's rewrite this equation as: x2 -19x +78= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +78x2 while the sum remains -19x
x2 - 13x  - 6x  +78= 0
=> x(x - 13) - 6 (x - 13) = 0
=> (x - 13) (x - 6) = 0
=>  (x - 13) =0  OR (x - 6) = 0
=> x = + 13  OR x = + 6
=> So, by factorization, we solve and find that the roots of the given equation are  + 13 , + 6

Example 84: Using the factorization method solve the quadratic equation: x2 -16x +63= 0

Let's rewrite this equation as: x2 -16x +63= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +63x2 while the sum remains -16x
x2 - 9x  - 7x  +63= 0
=> x(x - 9) - 7 (x - 9) = 0
=> (x - 9) (x - 7) = 0
=>  (x - 9) =0  OR (x - 7) = 0
=> x = + 9  OR x = + 7
=> So, by factorization, we solve and find that the roots of the given equation are  + 9 , + 7

Example 85: Using the factorization method solve the quadratic equation: x2 -7x -170= 0

Let's rewrite this equation as: x2 -7x -170= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is -170x2 while the sum remains -7x
x2 + 10x  - 17x  -170= 0
=> x(x + 10) - 17 (x + 10) = 0
=> (x + 10) (x - 17) = 0
=>  (x + 10) =0  OR (x - 17) = 0
=> x = - 10  OR x = + 17
=> So, by factorization, we solve and find that the roots of the given equation are  - 10 , + 17

Example 86: Using the factorization method solve the quadratic equation: x2 +18x -19= 0

Let's rewrite this equation as: x2 +18x -19= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is -19x2 while the sum remains +18x
x2 + 19x  - 1x  -19= 0
=> x(x + 19) - 1 (x + 19) = 0
=> (x + 19) (x - 1) = 0
=>  (x + 19) =0  OR (x - 1) = 0
=> x = - 19  OR x = + 1
=> So, by factorization, we solve and find that the roots of the given equation are  - 19 , + 1

Example 87: Using the factorization method solve the quadratic equation: x2 +7x +6= 0

Let's rewrite this equation as: x2 +7x +6= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +6x2 while the sum remains +7x
x2 + 1x  + 6x  +6= 0
=> x(x + 1) + 6 (x + 1) = 0
=> (x + 1) (x + 6) = 0
=>  (x + 1) =0  OR (x + 6) = 0
=> x = - 1  OR x = - 6
=> So, by factorization, we solve and find that the roots of the given equation are  - 1 , - 6

Example 88: Using the factorization method solve the quadratic equation: x2 +23x +130= 0

Let's rewrite this equation as: x2 +23x +130= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +130x2 while the sum remains +23x
x2 + 13x  + 10x  +130= 0
=> x(x + 13) + 10 (x + 13) = 0
=> (x + 13) (x + 10) = 0
=>  (x + 13) =0  OR (x + 10) = 0
=> x = - 13  OR x = - 10
=> So, by factorization, we solve and find that the roots of the given equation are  - 13 , - 10

Example 89: Using the factorization method solve the quadratic equation: x2 -17x +66= 0

Let's rewrite this equation as: x2 -17x +66= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +66x2 while the sum remains -17x
x2 - 11x  - 6x  +66= 0
=> x(x - 11) - 6 (x - 11) = 0
=> (x - 11) (x - 6) = 0
=>  (x - 11) =0  OR (x - 6) = 0
=> x = + 11  OR x = + 6
=> So, by factorization, we solve and find that the roots of the given equation are  + 11 , + 6

Example 90: Using the factorization method solve the quadratic equation: x2 +16x +39= 0

Let's rewrite this equation as: x2 +16x +39= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +39x2 while the sum remains +16x
x2 + 3x  + 13x  +39= 0
=> x(x + 3) + 13 (x + 3) = 0
=> (x + 3) (x + 13) = 0
=>  (x + 3) =0  OR (x + 13) = 0
=> x = - 3  OR x = - 13
=> So, by factorization, we solve and find that the roots of the given equation are  - 3 , - 13

Example 91: Using the factorization method solve the quadratic equation: x2 -15x +50= 0

Let's rewrite this equation as: x2 -15x +50= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +50x2 while the sum remains -15x
x2 - 10x  - 5x  +50= 0
=> x(x - 10) - 5 (x - 10) = 0
=> (x - 10) (x - 5) = 0
=>  (x - 10) =0  OR (x - 5) = 0
=> x = + 10  OR x = + 5
=> So, by factorization, we solve and find that the roots of the given equation are  + 10 , + 5

Example 92: Using the factorization method solve the quadratic equation: x2 +7x +12= 0

Let's rewrite this equation as: x2 +7x +12= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +12x2 while the sum remains +7x
x2 + 4x  + 3x  +12= 0
=> x(x + 4) + 3 (x + 4) = 0
=> (x + 4) (x + 3) = 0
=>  (x + 4) =0  OR (x + 3) = 0
=> x = - 4  OR x = - 3
=> So, by factorization, we solve and find that the roots of the given equation are  - 4 , - 3

Example 93: Using the factorization method solve the quadratic equation: x2 -23x -24= 0

Let's rewrite this equation as: x2 -23x -24= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is -24x2 while the sum remains -23x
x2 - 24x  + 1x  -24= 0
=> x(x - 24) + 1 (x - 24) = 0
=> (x - 24) (x + 1) = 0
=>  (x - 24) =0  OR (x + 1) = 0
=> x = + 24  OR x = - 1
=> So, by factorization, we solve and find that the roots of the given equation are  + 24 , - 1

Example 94: Using the factorization method solve the quadratic equation: x2 -24x +135= 0

Let's rewrite this equation as: x2 -24x +135= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +135x2 while the sum remains -24x
x2 - 9x  - 15x  +135= 0
=> x(x - 9) - 15 (x - 9) = 0
=> (x - 9) (x - 15) = 0
=>  (x - 9) =0  OR (x - 15) = 0
=> x = + 9  OR x = + 15
=> So, by factorization, we solve and find that the roots of the given equation are  + 9 , + 15

Example 95: Using the factorization method solve the quadratic equation: x2 -35x +300= 0

Let's rewrite this equation as: x2 -35x +300= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +300x2 while the sum remains -35x
x2 - 20x  - 15x  +300= 0
=> x(x - 20) - 15 (x - 20) = 0
=> (x - 20) (x - 15) = 0
=>  (x - 20) =0  OR (x - 15) = 0
=> x = + 20  OR x = + 15
=> So, by factorization, we solve and find that the roots of the given equation are  + 20 , + 15

Example 96: Using the factorization method solve the quadratic equation: x2 -40x +396= 0

Let's rewrite this equation as: x2 -40x +396= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +396x2 while the sum remains -40x
x2 - 22x  - 18x  +396= 0
=> x(x - 22) - 18 (x - 22) = 0
=> (x - 22) (x - 18) = 0
=>  (x - 22) =0  OR (x - 18) = 0
=> x = + 22  OR x = + 18
=> So, by factorization, we solve and find that the roots of the given equation are  + 22 , + 18

Example 97: Using the factorization method solve the quadratic equation: x2 -6x -391= 0

Let's rewrite this equation as: x2 -6x -391= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is -391x2 while the sum remains -6x
x2 + 17x  - 23x  -391= 0
=> x(x + 17) - 23 (x + 17) = 0
=> (x + 17) (x - 23) = 0
=>  (x + 17) =0  OR (x - 23) = 0
=> x = - 17  OR x = + 23
=> So, by factorization, we solve and find that the roots of the given equation are  - 17 , + 23

Example 98: Using the factorization method solve the quadratic equation: x2 +18x +17= 0

Let's rewrite this equation as: x2 +18x +17= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is +17x2 while the sum remains +18x
x2 + 1x  + 17x  +17= 0
=> x(x + 1) + 17 (x + 1) = 0
=> (x + 1) (x + 17) = 0
=>  (x + 1) =0  OR (x + 17) = 0
=> x = - 1  OR x = - 17
=> So, by factorization, we solve and find that the roots of the given equation are  - 1 , - 17

Example 99: Using the factorization method solve the quadratic equation: x2 -5x -14= 0

Let's rewrite this equation as: x2 -5x -14= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is -14x2 while the sum remains -5x
x2 + 2x  - 7x  -14= 0
=> x(x + 2) - 7 (x + 2) = 0
=> (x + 2) (x - 7) = 0
=>  (x + 2) =0  OR (x - 7) = 0
=> x = - 2  OR x = + 7
=> So, by factorization, we solve and find that the roots of the given equation are  - 2 , + 7

Example 100: Using the factorization method solve the quadratic equation: x2 -8x -345= 0

Let's rewrite this equation as: x2 -8x -345= 0
A key step here is to factorize the quadratic polynomial expression properly. The middle term needs to be split into two terms whose product is -345x2 while the sum remains -8x
x2 + 15x  - 23x  -345= 0
=> x(x + 15) - 23 (x + 15) = 0
=> (x + 15) (x - 23) = 0
=>  (x + 15) =0  OR (x - 23) = 0
=> x = - 15  OR x = + 23
=> So, by factorization, we solve and find that the roots of the given equation are  - 15 , + 23