H(t)=-32t^2+288t+1152

Solution for H(t)=-32t^2+288t+1152 equation:

(H)=-32H^2+288H+1152

We move all terms to the left:

(H)-(-32H^2+288H+1152)=0

We get rid of parentheses

32H^2-288H+H-1152=0

We add all the numbers together, and all the variables

32H^2-287H-1152=0

a = 32; b = -287; c = -1152;
Δ = b2-4ac
Δ = -2872-4·32·(-1152)
Δ = 229825
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:

$\sqrt{\Delta}=\sqrt{229825}=\sqrt{25*9193}=\sqrt{25}*\sqrt{9193}=5\sqrt{9193}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-287)-5\sqrt{9193}}{2*32}=\frac{287-5\sqrt{9193}}{64}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-287)+5\sqrt{9193}}{2*32}=\frac{287+5\sqrt{9193}}{64}
$