H(t)=-16t^2+25t+80

Solution for H(t)=-16t^2+25t+80 equation:

(H)=-16H^2+25H+80

We move all terms to the left:

(H)-(-16H^2+25H+80)=0

We get rid of parentheses

16H^2-25H+H-80=0

We add all the numbers together, and all the variables

16H^2-24H-80=0

a = 16; b = -24; c = -80;
Δ = b2-4ac
Δ = -242-4·16·(-80)
Δ = 5696
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{5696}=\sqrt{64*89}=\sqrt{64}*\sqrt{89}=8\sqrt{89}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-8\sqrt{89}}{2*16}=\frac{24-8\sqrt{89}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+8\sqrt{89}}{2*16}=\frac{24+8\sqrt{89}}{32}
$