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

Solution for H(2)=-16t^2+48t equation:

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

We move all terms to the left:

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

We get rid of parentheses

16H^2-48H+2=0

a = 16; b = -48; c = +2;
Δ = b2-4ac
Δ = -482-4·16·2
Δ = 2176
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:

$\sqrt{\Delta}=\sqrt{2176}=\sqrt{64*34}=\sqrt{64}*\sqrt{34}=8\sqrt{34}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-8\sqrt{34}}{2*16}=\frac{48-8\sqrt{34}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+8\sqrt{34}}{2*16}=\frac{48+8\sqrt{34}}{32}
$