H=-16t^2+32+20

Solution for H=-16t^2+32+20 equation:

=-16H^2+32+20

We move all terms to the left:

-(-16H^2+32+20)=0

We get rid of parentheses

16H^2-32-20=0

We add all the numbers together, and all the variables

16H^2-52=0

a = 16; b = 0; c = -52;
Δ = b2-4ac
Δ = 02-4·16·(-52)
Δ = 3328
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:

$\sqrt{\Delta}=\sqrt{3328}=\sqrt{256*13}=\sqrt{256}*\sqrt{13}=16\sqrt{13}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{13}}{2*16}=\frac{0-16\sqrt{13}}{32}
=-\frac{16\sqrt{13}}{32}
=-\frac{\sqrt{13}}{2}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{13}}{2*16}=\frac{0+16\sqrt{13}}{32}
=\frac{16\sqrt{13}}{32}
=\frac{\sqrt{13}}{2}
$