z^2=216

Solution for z^2=216 equation:

z^2=216

We move all terms to the left:

z^2-(216)=0

a = 1; b = 0; c = -216;
Δ = b2-4ac
Δ = 02-4·1·(-216)
Δ = 864
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{864}=\sqrt{144*6}=\sqrt{144}*\sqrt{6}=12\sqrt{6}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{6}}{2*1}=\frac{0-12\sqrt{6}}{2}
=-\frac{12\sqrt{6}}{2}
=-6\sqrt{6}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{6}}{2*1}=\frac{0+12\sqrt{6}}{2}
=\frac{12\sqrt{6}}{2}
=6\sqrt{6}
$