H=48t-2t^2

Solution for H=48t-2t^2 equation:

=48H-2H^2

We move all terms to the left:

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

We get rid of parentheses

2H^2-48H=0

a = 2; b = -48; c = 0;
Δ = b2-4ac
Δ = -482-4·2·0
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*2}=\frac{0}{4}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+48}{2*2}=\frac{96}{4}
=24
$