3k^2-4K+4=96

Solution for 3k^2-4K+4=96 equation:

3k^2-4+4=96

We move all terms to the left:

3k^2-4+4-(96)=0

We add all the numbers together, and all the variables

3k^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:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

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