H=-16t^2+124t+32

Solution for H=-16t^2+124t+32 equation:

=-16H^2+124H+32

We move all terms to the left:

-(-16H^2+124H+32)=0

We get rid of parentheses

16H^2-124H-32=0

a = 16; b = -124; c = -32;
Δ = b2-4ac
Δ = -1242-4·16·(-32)
Δ = 17424
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{17424}=132$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-124)-132}{2*16}=\frac{-8}{32}
=-1/4
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-124)+132}{2*16}=\frac{256}{32}
=8
$