S=2h2+2h

Solution for S=2h2+2h equation:

=2S^2+2S

We move all terms to the left:

-(2S^2+2S)=0

We get rid of parentheses

-2S^2-2S=0

a = -2; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·(-2)·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*-2}=\frac{0}{-4}
=0
$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*-2}=\frac{4}{-4}
=-1
$