H(t)=-16t^2+4t+80

Solution for H(t)=-16t^2+4t+80 equation:

(H)=-16H^2+4H+80

We move all terms to the left:

(H)-(-16H^2+4H+80)=0

We get rid of parentheses

16H^2-4H+H-80=0

We add all the numbers together, and all the variables

16H^2-3H-80=0

a = 16; b = -3; c = -80;
Δ = b2-4ac
Δ = -32-4·16·(-80)
Δ = 5129
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{-(-3)-\sqrt{5129}}{2*16}=\frac{3-\sqrt{5129}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{5129}}{2*16}=\frac{3+\sqrt{5129}}{32}
$