-2h2+40h=-62

Solution for -2h2+40h=-62 equation:

-2h^2+40h=-62

We move all terms to the left:

-2h^2+40h-(-62)=0

We add all the numbers together, and all the variables

-2h^2+40h+62=0

a = -2; b = 40; c = +62;
Δ = b2-4ac
Δ = 402-4·(-2)·62
Δ = 2096
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:

$\sqrt{\Delta}=\sqrt{2096}=\sqrt{16*131}=\sqrt{16}*\sqrt{131}=4\sqrt{131}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-4\sqrt{131}}{2*-2}=\frac{-40-4\sqrt{131}}{-4}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+4\sqrt{131}}{2*-2}=\frac{-40+4\sqrt{131}}{-4}
$