H=-16t^2+128

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

=-16H^2+128

We move all terms to the left:

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

We get rid of parentheses

16H^2-128=0

a = 16; b = 0; c = -128;
Δ = b2-4ac
Δ = 02-4·16·(-128)
Δ = 8192
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:

$\sqrt{\Delta}=\sqrt{8192}=\sqrt{4096*2}=\sqrt{4096}*\sqrt{2}=64\sqrt{2}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-64\sqrt{2}}{2*16}=\frac{0-64\sqrt{2}}{32}
=-\frac{64\sqrt{2}}{32}
=-2\sqrt{2}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+64\sqrt{2}}{2*16}=\frac{0+64\sqrt{2}}{32}
=\frac{64\sqrt{2}}{32}
=2\sqrt{2}
$