2x^2+18x-20=0
I will assume you are solving.
divided each term by 2
x^2 + 9x - 10 = 0
(x+10)(x-1) = 0
x = -10 or x = 1
To solve the quadratic equation 2x^2 + 18x - 20 = 0, we can use the quadratic formula, which is:
x = (-b ± √(b^2 - 4ac)) / 2a
In the given equation, a = 2, b = 18, and c = -20. Substituting these values into the quadratic formula, we get:
x = (-(18) ± √((18)^2 - 4(2)(-20))) / 2(2)
Simplifying further:
x = (-18 ± √(324 + 160)) / 4
= (-18 ± √484) / 4
= (-18 ± 22) / 4
There are two possible solutions:
1. When we take the + (plus) in the ± symbol:
x = (-18 + 22) / 4 = 4 / 4 = 1
2. When we take the - (minus) in the ± symbol:
x = (-18 - 22) / 4 = -40 / 4 = -10
Therefore, the solutions to the quadratic equation 2x^2 + 18x - 20 = 0 are x = 1 and x = -10.