H=-10t2+25t

Solution for H=-10t2+25t equation:

=-10H^2+25H

We move all terms to the left:

-(-10H^2+25H)=0

We get rid of parentheses

10H^2-25H=0

a = 10; b = -25; c = 0;
Δ = b2-4ac
Δ = -252-4·10·0
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-25}{2*10}=\frac{0}{20}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+25}{2*10}=\frac{50}{20}
=2+1/2
$