-96x^2+72x-12=0

Solution for -96x^2+72x-12=0 equation:

-96x^2+72x-12=0

a = -96; b = 72; c = -12;
Δ = b2-4ac
Δ = 722-4·(-96)·(-12)
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(72)-24}{2*-96}=\frac{-96}{-192}
=1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(72)+24}{2*-96}=\frac{-48}{-192}
=1/4
$