H=16t^2+180t

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

=16H^2+180H

We move all terms to the left:

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

We get rid of parentheses

-16H^2-180H=0

a = -16; b = -180; c = 0;
Δ = b2-4ac
Δ = -1802-4·(-16)·0
Δ = 32400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{32400}=180$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-180)-180}{2*-16}=\frac{0}{-32}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-180)+180}{2*-16}=\frac{360}{-32}
=-11+1/4
$