H=-16t^2-20t+220

Solution for H=-16t^2-20t+220 equation:

=-16H^2-20H+220

We move all terms to the left:

-(-16H^2-20H+220)=0

We get rid of parentheses

16H^2+20H-220=0

a = 16; b = 20; c = -220;
Δ = b2-4ac
Δ = 202-4·16·(-220)
Δ = 14480
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:

$\sqrt{\Delta}=\sqrt{14480}=\sqrt{16*905}=\sqrt{16}*\sqrt{905}=4\sqrt{905}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{905}}{2*16}=\frac{-20-4\sqrt{905}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{905}}{2*16}=\frac{-20+4\sqrt{905}}{32}
$