H=-6t^2+96

Solution for H=-6t^2+96 equation:

=-6H^2+96

We move all terms to the left:

-(-6H^2+96)=0

We get rid of parentheses

6H^2-96=0

a = 6; b = 0; c = -96;
Δ = b2-4ac
Δ = 02-4·6·(-96)
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-48}{2*6}=\frac{-48}{12}
=-4
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+48}{2*6}=\frac{48}{12}
=4
$