-80x+12x^2+96=0

Solution for -80x+12x^2+96=0 equation:

-80x+12x^2+96=0

a = 12; b = -80; c = +96;
Δ = b2-4ac
Δ = -802-4·12·96
Δ = 1792
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:

$\sqrt{\Delta}=\sqrt{1792}=\sqrt{256*7}=\sqrt{256}*\sqrt{7}=16\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-16\sqrt{7}}{2*12}=\frac{80-16\sqrt{7}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+16\sqrt{7}}{2*12}=\frac{80+16\sqrt{7}}{24}
$