H=-16t2+95t

Solution for H=-16t2+95t equation:

=-16H^2+95H

We move all terms to the left:

-(-16H^2+95H)=0

We get rid of parentheses

16H^2-95H=0

a = 16; b = -95; c = 0;
Δ = b2-4ac
Δ = -952-4·16·0
Δ = 9025
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9025}=95$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-95)-95}{2*16}=\frac{0}{32}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-95)+95}{2*16}=\frac{190}{32}
=5+15/16
$