H=-16t^2+20t+8

Solution for H=-16t^2+20t+8 equation:

=-16H^2+20H+8

We move all terms to the left:

-(-16H^2+20H+8)=0

We get rid of parentheses

16H^2-20H-8=0

a = 16; b = -20; c = -8;
Δ = b2-4ac
Δ = -202-4·16·(-8)
Δ = 912
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:

$\sqrt{\Delta}=\sqrt{912}=\sqrt{16*57}=\sqrt{16}*\sqrt{57}=4\sqrt{57}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{57}}{2*16}=\frac{20-4\sqrt{57}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{57}}{2*16}=\frac{20+4\sqrt{57}}{32}
$