500=120x-4x^2

Solution for 500=120x-4x^2 equation:

500=120x-4x^2

We move all terms to the left:

500-(120x-4x^2)=0

We get rid of parentheses

4x^2-120x+500=0

a = 4; b = -120; c = +500;
Δ = b2-4ac
Δ = -1202-4·4·500
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-120)-80}{2*4}=\frac{40}{8}
=5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-120)+80}{2*4}=\frac{200}{8}
=25
$