H=-5.t2+20.t

Solution for H=-5.t2+20.t equation:

=-5.H^2+20.H

We move all terms to the left:

-(-5.H^2+20.H)=0

We get rid of parentheses

5.H^2-20.H=0

We add all the numbers together, and all the variables

5.H^2-20H=0

a = 5.; b = -20; c = 0;
Δ = b2-4ac
Δ = -202-4·5.·0
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-20}{2*5.}=\frac{0}{10}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+20}{2*5.}=\frac{40}{10}
=4
$