-3x^2=-96

Solution for -3x^2=-96 equation:

-3x^2=-96

We move all terms to the left:

-3x^2-(-96)=0

We add all the numbers together, and all the variables

-3x^2+96=0

a = -3; b = 0; c = +96;
Δ = b2-4ac
Δ = 02-4·(-3)·96
Δ = 1152
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:

$\sqrt{\Delta}=\sqrt{1152}=\sqrt{576*2}=\sqrt{576}*\sqrt{2}=24\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24\sqrt{2}}{2*-3}=\frac{0-24\sqrt{2}}{-6}
=-\frac{24\sqrt{2}}{-6}
=-\frac{4\sqrt{2}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24\sqrt{2}}{2*-3}=\frac{0+24\sqrt{2}}{-6}
=\frac{24\sqrt{2}}{-6}
=\frac{4\sqrt{2}}{-1}
$