H(t)=-16t2+64t+80

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

16H^2-63H-80=0

a = 16; b = -63; c = -80;
Δ = b2-4ac
Δ = -632-4·16·(-80)
Δ = 9089
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{-(-63)-\sqrt{9089}}{2*16}=\frac{63-\sqrt{9089}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-63)+\sqrt{9089}}{2*16}=\frac{63+\sqrt{9089}}{32}
$