-16x^2+96x-12=120

Solution for -16x^2+96x-12=120 equation:

-16x^2+96x-12=120

We move all terms to the left:

-16x^2+96x-12-(120)=0

We add all the numbers together, and all the variables

-16x^2+96x-132=0

a = -16; b = 96; c = -132;
Δ = b2-4ac
Δ = 962-4·(-16)·(-132)
Δ = 768
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:

$\sqrt{\Delta}=\sqrt{768}=\sqrt{256*3}=\sqrt{256}*\sqrt{3}=16\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-16\sqrt{3}}{2*-16}=\frac{-96-16\sqrt{3}}{-32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+16\sqrt{3}}{2*-16}=\frac{-96+16\sqrt{3}}{-32}
$