Question

Solve for x by completing the square: -2x^2+2x+10=0

Answer 1

\[\LARGE x_{1/2}={-b\pm\sqrt{b^2-4ac}\over 2a}\]
\[\LARGE ax^2+bx+c=0\]
in your case you have:
\[\LARGE \underbrace{-2}_{a}\cdot x^2\;\;\;\;\underbrace{+2}_{b}\cdot x\;\;\;\; \underbrace{+10}_{c}=0\]

Answer 2

now substitute...

Answer 3

-2x^2+2x+10=0 ---> devide both sides by -2

x^2 - x - 5 = 0 ----> transpose 5 to other side.

x^2 -x = 5 ----> divide the only x term by 2 and square it.

x^2 - (x/2)^2 = 5 -----> know add the answer you got from the step above to both sides which is 1/4.

x^2 + x + 1/4 = 5 + 1/4 ----> the square is competed just simplify it.

(x+1/2)^2 = 21/4 ----> simplify again.

x + 1/2 = (sqrt(7))/2

x = [(sqrt(7))/2] - 1/2

x = ([sqrt(7)] - 1)/2