H=-16t^2+128t+1.5

Solution for H=-16t^2+128t+1.5 equation:

=-16H^2+128H+1.5

We move all terms to the left:

-(-16H^2+128H+1.5)=0

We get rid of parentheses

16H^2-128H-1.5=0

a = 16; b = -128; c = -1.5;
Δ = b2-4ac
Δ = -1282-4·16·(-1.5)
Δ = 16480
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:

$\sqrt{\Delta}=\sqrt{16480}=\sqrt{16*1030}=\sqrt{16}*\sqrt{1030}=4\sqrt{1030}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-128)-4\sqrt{1030}}{2*16}=\frac{128-4\sqrt{1030}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-128)+4\sqrt{1030}}{2*16}=\frac{128+4\sqrt{1030}}{32}
$