H=16t^2+95t

Solution for H=16t^2+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
$