80=2x^2+2x+40

Solution for 80=2x^2+2x+40 equation:

80=2x^2+2x+40

We move all terms to the left:

80-(2x^2+2x+40)=0

We get rid of parentheses

-2x^2-2x-40+80=0

We add all the numbers together, and all the variables

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