0=-2x^2-2x+40

Solution for 0=-2x^2-2x+40 equation:

0=-2x^2-2x+40

We move all terms to the left:

0-(-2x^2-2x+40)=0

We add all the numbers together, and all the variables

-(-2x^2-2x+40)=0

We get rid of parentheses

2x^2+2x-40=0

a = 2; b = 2; c = -40;
Δ = b2-4ac
Δ = 22-4·2·(-40)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-18}{2*2}=\frac{-20}{4}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+18}{2*2}=\frac{16}{4}
=4
$