0=-16x^2+160x+120

Solution for 0=-16x^2+160x+120 equation:

0=-16x^2+160x+120

We move all terms to the left:

0-(-16x^2+160x+120)=0

We add all the numbers together, and all the variables

-(-16x^2+160x+120)=0

We get rid of parentheses

16x^2-160x-120=0

a = 16; b = -160; c = -120;
Δ = b2-4ac
Δ = -1602-4·16·(-120)
Δ = 33280
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:

$\sqrt{\Delta}=\sqrt{33280}=\sqrt{256*130}=\sqrt{256}*\sqrt{130}=16\sqrt{130}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-160)-16\sqrt{130}}{2*16}=\frac{160-16\sqrt{130}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-160)+16\sqrt{130}}{2*16}=\frac{160+16\sqrt{130}}{32}
$