3k^2=162

Solution for 3k^2=162 equation:

3k^2=162

We move all terms to the left:

3k^2-(162)=0

a = 3; b = 0; c = -162;
Δ = b2-4ac
Δ = 02-4·3·(-162)
Δ = 1944
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:

$\sqrt{\Delta}=\sqrt{1944}=\sqrt{324*6}=\sqrt{324}*\sqrt{6}=18\sqrt{6}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-18\sqrt{6}}{2*3}=\frac{0-18\sqrt{6}}{6}
=-\frac{18\sqrt{6}}{6}
=-3\sqrt{6}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+18\sqrt{6}}{2*3}=\frac{0+18\sqrt{6}}{6}
=\frac{18\sqrt{6}}{6}
=3\sqrt{6}
$