H=-16t^2+120t+300

Solution for H=-16t^2+120t+300 equation:

=-16H^2+120H+300

We move all terms to the left:

-(-16H^2+120H+300)=0

We get rid of parentheses

16H^2-120H-300=0

a = 16; b = -120; c = -300;
Δ = b2-4ac
Δ = -1202-4·16·(-300)
Δ = 33600
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:

$\sqrt{\Delta}=\sqrt{33600}=\sqrt{1600*21}=\sqrt{1600}*\sqrt{21}=40\sqrt{21}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-120)-40\sqrt{21}}{2*16}=\frac{120-40\sqrt{21}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-120)+40\sqrt{21}}{2*16}=\frac{120+40\sqrt{21}}{32}
$