H=-16t^2+128t+48

Solution for H=-16t^2+128t+48 equation:

=-16H^2+128H+48

We move all terms to the left:

-(-16H^2+128H+48)=0

We get rid of parentheses

16H^2-128H-48=0

a = 16; b = -128; c = -48;
Δ = b2-4ac
Δ = -1282-4·16·(-48)
Δ = 19456
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:

$\sqrt{\Delta}=\sqrt{19456}=\sqrt{1024*19}=\sqrt{1024}*\sqrt{19}=32\sqrt{19}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-128)-32\sqrt{19}}{2*16}=\frac{128-32\sqrt{19}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-128)+32\sqrt{19}}{2*16}=\frac{128+32\sqrt{19}}{32}
$