H=-16t^2+95t+5

Solution for H=-16t^2+95t+5 equation:

=-16H^2+95H+5

We move all terms to the left:

-(-16H^2+95H+5)=0

We get rid of parentheses

16H^2-95H-5=0

a = 16; b = -95; c = -5;
Δ = b2-4ac
Δ = -952-4·16·(-5)
Δ = 9345
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-95)-\sqrt{9345}}{2*16}=\frac{95-\sqrt{9345}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-95)+\sqrt{9345}}{2*16}=\frac{95+\sqrt{9345}}{32}
$