H(t)=-6t^2+48t

Solution for H(t)=-6t^2+48t equation:

(H)=-6H^2+48H

We move all terms to the left:

(H)-(-6H^2+48H)=0

We get rid of parentheses

6H^2-48H+H=0

We add all the numbers together, and all the variables

6H^2-47H=0

a = 6; b = -47; c = 0;
Δ = b2-4ac
Δ = -472-4·6·0
Δ = 2209
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2209}=47$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-47)-47}{2*6}=\frac{0}{12}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-47)+47}{2*6}=\frac{94}{12}
=7+5/6
$