12=6z2

Solution for 12=6z2 equation:

12=6z^2

We move all terms to the left:

12-(6z^2)=0

a = -6; b = 0; c = +12;
Δ = b2-4ac
Δ = 02-4·(-6)·12
Δ = 288
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:

$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{2}}{2*-6}=\frac{0-12\sqrt{2}}{-12}
=-\frac{12\sqrt{2}}{-12}
=-\frac{\sqrt{2}}{-1}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{2}}{2*-6}=\frac{0+12\sqrt{2}}{-12}
=\frac{12\sqrt{2}}{-12}
=\frac{\sqrt{2}}{-1}
$