H=-16t2+64t+80

Solution for H=-16t2+64t+80 equation:

=-16H^2+64H+80

We move all terms to the left:

-(-16H^2+64H+80)=0

We get rid of parentheses

16H^2-64H-80=0

a = 16; b = -64; c = -80;
Δ = b2-4ac
Δ = -642-4·16·(-80)
Δ = 9216
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9216}=96$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-64)-96}{2*16}=\frac{-32}{32}
=-1
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-64)+96}{2*16}=\frac{160}{32}
=5
$