12=2w^2

Solution for 12=2w^2 equation:

12=2w^2

We move all terms to the left:

12-(2w^2)=0

a = -2; b = 0; c = +12;
Δ = b2-4ac
Δ = 02-4·(-2)·12
Δ = 96
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{6}}{2*-2}=\frac{0-4\sqrt{6}}{-4}
=-\frac{4\sqrt{6}}{-4}
=-\frac{\sqrt{6}}{-1}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{6}}{2*-2}=\frac{0+4\sqrt{6}}{-4}
=\frac{4\sqrt{6}}{-4}
=\frac{\sqrt{6}}{-1}
$