H=-16t2+48t

Solution for H=-16t2+48t equation:

=-16H^2+48H

We move all terms to the left:

-(-16H^2+48H)=0

We get rid of parentheses

16H^2-48H=0

a = 16; b = -48; c = 0;
Δ = b2-4ac
Δ = -482-4·16·0
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{-(-48)-48}{2*16}=\frac{0}{32}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+48}{2*16}=\frac{96}{32}
=3
$