H=-16t^2+65t

Solution for H=-16t^2+65t equation:

=-16H^2+65H

We move all terms to the left:

-(-16H^2+65H)=0

We get rid of parentheses

16H^2-65H=0

a = 16; b = -65; c = 0;
Δ = b2-4ac
Δ = -652-4·16·0
Δ = 4225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4225}=65$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-65)-65}{2*16}=\frac{0}{32}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-65)+65}{2*16}=\frac{130}{32}
=4+1/16
$