H=-2t^2+10t

Solution for H=-2t^2+10t equation:

=-2H^2+10H

We move all terms to the left:

-(-2H^2+10H)=0

We get rid of parentheses

2H^2-10H=0

a = 2; b = -10; c = 0;
Δ = b2-4ac
Δ = -102-4·2·0
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-10}{2*2}=\frac{0}{4}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+10}{2*2}=\frac{20}{4}
=5
$