k^2=54

Solution for k^2=54 equation:

k^2=54

We move all terms to the left:

k^2-(54)=0

a = 1; b = 0; c = -54;
Δ = b2-4ac
Δ = 02-4·1·(-54)
Δ = 216
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:

$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{6}}{2*1}=\frac{0-6\sqrt{6}}{2}
=-\frac{6\sqrt{6}}{2}
=-3\sqrt{6}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{6}}{2*1}=\frac{0+6\sqrt{6}}{2}
=\frac{6\sqrt{6}}{2}
=3\sqrt{6}
$