-3x^2+24x-12=0

Solution for -3x^2+24x-12=0 equation:

-3x^2+24x-12=0

a = -3; b = 24; c = -12;
Δ = b2-4ac
Δ = 242-4·(-3)·(-12)
Δ = 432
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:

$\sqrt{\Delta}=\sqrt{432}=\sqrt{144*3}=\sqrt{144}*\sqrt{3}=12\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-12\sqrt{3}}{2*-3}=\frac{-24-12\sqrt{3}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+12\sqrt{3}}{2*-3}=\frac{-24+12\sqrt{3}}{-6}
$