H=-16t^2+40t+108

Solution for H=-16t^2+40t+108 equation:

=-16H^2+40H+108

We move all terms to the left:

-(-16H^2+40H+108)=0

We get rid of parentheses

16H^2-40H-108=0

a = 16; b = -40; c = -108;
Δ = b2-4ac
Δ = -402-4·16·(-108)
Δ = 8512
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:

$\sqrt{\Delta}=\sqrt{8512}=\sqrt{64*133}=\sqrt{64}*\sqrt{133}=8\sqrt{133}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-8\sqrt{133}}{2*16}=\frac{40-8\sqrt{133}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+8\sqrt{133}}{2*16}=\frac{40+8\sqrt{133}}{32}
$