200=-16x^2+120x

Solution for 200=-16x^2+120x equation:

200=-16x^2+120x

We move all terms to the left:

200-(-16x^2+120x)=0

We get rid of parentheses

16x^2-120x+200=0

a = 16; b = -120; c = +200;
Δ = b2-4ac
Δ = -1202-4·16·200
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-120)-40}{2*16}=\frac{80}{32}
=2+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-120)+40}{2*16}=\frac{160}{32}
=5
$