6k^2=312

Solution for 6k^2=312 equation:

6k^2=312

We move all terms to the left:

6k^2-(312)=0

a = 6; b = 0; c = -312;
Δ = b2-4ac
Δ = 02-4·6·(-312)
Δ = 7488
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:

$\sqrt{\Delta}=\sqrt{7488}=\sqrt{576*13}=\sqrt{576}*\sqrt{13}=24\sqrt{13}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24\sqrt{13}}{2*6}=\frac{0-24\sqrt{13}}{12}
=-\frac{24\sqrt{13}}{12}
=-2\sqrt{13}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24\sqrt{13}}{2*6}=\frac{0+24\sqrt{13}}{12}
=\frac{24\sqrt{13}}{12}
=2\sqrt{13}
$