0=-16x^2+120x+4

Solution for 0=-16x^2+120x+4 equation:

0=-16x^2+120x+4

We move all terms to the left:

0-(-16x^2+120x+4)=0

We add all the numbers together, and all the variables

-(-16x^2+120x+4)=0

We get rid of parentheses

16x^2-120x-4=0

a = 16; b = -120; c = -4;
Δ = b2-4ac
Δ = -1202-4·16·(-4)
Δ = 14656
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:

$\sqrt{\Delta}=\sqrt{14656}=\sqrt{64*229}=\sqrt{64}*\sqrt{229}=8\sqrt{229}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-120)-8\sqrt{229}}{2*16}=\frac{120-8\sqrt{229}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-120)+8\sqrt{229}}{2*16}=\frac{120+8\sqrt{229}}{32}
$