H=-16t^2+48t+100

Solution for H=-16t^2+48t+100 equation:

=-16H^2+48H+100

We move all terms to the left:

-(-16H^2+48H+100)=0

We get rid of parentheses

16H^2-48H-100=0

a = 16; b = -48; c = -100;
Δ = b2-4ac
Δ = -482-4·16·(-100)
Δ = 8704
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:

$\sqrt{\Delta}=\sqrt{8704}=\sqrt{256*34}=\sqrt{256}*\sqrt{34}=16\sqrt{34}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-16\sqrt{34}}{2*16}=\frac{48-16\sqrt{34}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+16\sqrt{34}}{2*16}=\frac{48+16\sqrt{34}}{32}
$