H=-4t^2+40t

Solution for H=-4t^2+40t equation:

=-4H^2+40H

We move all terms to the left:

-(-4H^2+40H)=0

We get rid of parentheses

4H^2-40H=0

a = 4; b = -40; c = 0;
Δ = b2-4ac
Δ = -402-4·4·0
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-40}{2*4}=\frac{0}{8}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+40}{2*4}=\frac{80}{8}
=10
$