H=3+180t-16t^2

Solution for H=3+180t-16t^2 equation:

=3+180H-16H^2

We move all terms to the left:

-(3+180H-16H^2)=0

We get rid of parentheses

16H^2-180H-3=0

a = 16; b = -180; c = -3;
Δ = b2-4ac
Δ = -1802-4·16·(-3)
Δ = 32592
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:

$\sqrt{\Delta}=\sqrt{32592}=\sqrt{16*2037}=\sqrt{16}*\sqrt{2037}=4\sqrt{2037}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-180)-4\sqrt{2037}}{2*16}=\frac{180-4\sqrt{2037}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-180)+4\sqrt{2037}}{2*16}=\frac{180+4\sqrt{2037}}{32}
$