180=-16x^2+120x

Solution for 180=-16x^2+120x equation:

180=-16x^2+120x

We move all terms to the left:

180-(-16x^2+120x)=0

We get rid of parentheses

16x^2-120x+180=0

a = 16; b = -120; c = +180;
Δ = b2-4ac
Δ = -1202-4·16·180
Δ = 2880
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2880}=\sqrt{576*5}=\sqrt{576}*\sqrt{5}=24\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-120)-24\sqrt{5}}{2*16}=\frac{120-24\sqrt{5}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-120)+24\sqrt{5}}{2*16}=\frac{120+24\sqrt{5}}{32}
$