H=-16t2+80

Solution for H=-16t2+80 equation:

=-16H^2+80

We move all terms to the left:

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

We get rid of parentheses

16H^2-80=0

a = 16; b = 0; c = -80;
Δ = b2-4ac
Δ = 02-4·16·(-80)
Δ = 5120
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:

$\sqrt{\Delta}=\sqrt{5120}=\sqrt{1024*5}=\sqrt{1024}*\sqrt{5}=32\sqrt{5}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32\sqrt{5}}{2*16}=\frac{0-32\sqrt{5}}{32}
=-\frac{32\sqrt{5}}{32}
=-\sqrt{5}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32\sqrt{5}}{2*16}=\frac{0+32\sqrt{5}}{32}
=\frac{32\sqrt{5}}{32}
=\sqrt{5}
$