H(T)=-16t^2+48t+56

Solution for H(T)=-16t^2+48t+56 equation:

()=-16H^2+48H+56

We move all terms to the left:

()-(-16H^2+48H+56)=0

We add all the numbers together, and all the variables

-(-16H^2+48H+56)=0

We get rid of parentheses

16H^2-48H-56=0

a = 16; b = -48; c = -56;
Δ = b2-4ac
Δ = -482-4·16·(-56)
Δ = 5888
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:

$\sqrt{\Delta}=\sqrt{5888}=\sqrt{256*23}=\sqrt{256}*\sqrt{23}=16\sqrt{23}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-16\sqrt{23}}{2*16}=\frac{48-16\sqrt{23}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+16\sqrt{23}}{2*16}=\frac{48+16\sqrt{23}}{32}
$