H=-16t^2+80t-1

Solution for H=-16t^2+80t-1 equation:

=-16H^2+80H-1

We move all terms to the left:

-(-16H^2+80H-1)=0

We get rid of parentheses

16H^2-80H+1=0

a = 16; b = -80; c = +1;
Δ = b2-4ac
Δ = -802-4·16·1
Δ = 6336
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:

$\sqrt{\Delta}=\sqrt{6336}=\sqrt{576*11}=\sqrt{576}*\sqrt{11}=24\sqrt{11}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-24\sqrt{11}}{2*16}=\frac{80-24\sqrt{11}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+24\sqrt{11}}{2*16}=\frac{80+24\sqrt{11}}{32}
$