20=2h^2+10h

Solution for 20=2h^2+10h equation:

20=2h^2+10h

We move all terms to the left:

20-(2h^2+10h)=0

We get rid of parentheses

-2h^2-10h+20=0

a = -2; b = -10; c = +20;
Δ = b2-4ac
Δ = -102-4·(-2)·20
Δ = 260
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{260}=\sqrt{4*65}=\sqrt{4}*\sqrt{65}=2\sqrt{65}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{65}}{2*-2}=\frac{10-2\sqrt{65}}{-4}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{65}}{2*-2}=\frac{10+2\sqrt{65}}{-4}
$