20=2x^2+2x

Solution for 20=2x^2+2x equation:

20=2x^2+2x

We move all terms to the left:

20-(2x^2+2x)=0

We get rid of parentheses

-2x^2-2x+20=0

a = -2; b = -2; c = +20;
Δ = b2-4ac
Δ = -22-4·(-2)·20
Δ = 164
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{164}=\sqrt{4*41}=\sqrt{4}*\sqrt{41}=2\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{41}}{2*-2}=\frac{2-2\sqrt{41}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{41}}{2*-2}=\frac{2+2\sqrt{41}}{-4}
$