0=-16x^2+120x

Solution for 0=-16x^2+120x equation:

0=-16x^2+120x

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

16x^2-120x=0

a = 16; b = -120; c = 0;
Δ = b2-4ac
Δ = -1202-4·16·0
Δ = 14400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{14400}=120$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-120)-120}{2*16}=\frac{0}{32}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-120)+120}{2*16}=\frac{240}{32}
=7+1/2
$