0=-16x^2+120x+8

Solution for 0=-16x^2+120x+8 equation:

0=-16x^2+120x+8

We move all terms to the left:

0-(-16x^2+120x+8)=0

We add all the numbers together, and all the variables

-(-16x^2+120x+8)=0

We get rid of parentheses

16x^2-120x-8=0

a = 16; b = -120; c = -8;
Δ = b2-4ac
Δ = -1202-4·16·(-8)
Δ = 14912
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:

$\sqrt{\Delta}=\sqrt{14912}=\sqrt{64*233}=\sqrt{64}*\sqrt{233}=8\sqrt{233}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-120)-8\sqrt{233}}{2*16}=\frac{120-8\sqrt{233}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-120)+8\sqrt{233}}{2*16}=\frac{120+8\sqrt{233}}{32}
$