H=-16t^2-4+412

Solution for H=-16t^2-4+412 equation:

=-16H^2-4+412

We move all terms to the left:

-(-16H^2-4+412)=0

We get rid of parentheses

16H^2+4-412=0

We add all the numbers together, and all the variables

16H^2-408=0

a = 16; b = 0; c = -408;
Δ = b2-4ac
Δ = 02-4·16·(-408)
Δ = 26112
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:

$\sqrt{\Delta}=\sqrt{26112}=\sqrt{256*102}=\sqrt{256}*\sqrt{102}=16\sqrt{102}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{102}}{2*16}=\frac{0-16\sqrt{102}}{32}
=-\frac{16\sqrt{102}}{32}
=-\frac{\sqrt{102}}{2}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{102}}{2*16}=\frac{0+16\sqrt{102}}{32}
=\frac{16\sqrt{102}}{32}
=\frac{\sqrt{102}}{2}
$