H=-16t^2+32+3

Solution for H=-16t^2+32+3 equation:

=-16H^2+32+3

We move all terms to the left:

-(-16H^2+32+3)=0

We get rid of parentheses

16H^2-32-3=0

We add all the numbers together, and all the variables

16H^2-35=0

a = 16; b = 0; c = -35;
Δ = b2-4ac
Δ = 02-4·16·(-35)
Δ = 2240
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:

$\sqrt{\Delta}=\sqrt{2240}=\sqrt{64*35}=\sqrt{64}*\sqrt{35}=8\sqrt{35}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{35}}{2*16}=\frac{0-8\sqrt{35}}{32}
=-\frac{8\sqrt{35}}{32}
=-\frac{\sqrt{35}}{4}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{35}}{2*16}=\frac{0+8\sqrt{35}}{32}
=\frac{8\sqrt{35}}{32}
=\frac{\sqrt{35}}{4}
$