0=2w^2-96

Solution for 0=2w^2-96 equation:

0=2w^2-96

We move all terms to the left:

0-(2w^2-96)=0

We add all the numbers together, and all the variables

-(2w^2-96)=0

We get rid of parentheses

-2w^2+96=0

a = -2; b = 0; c = +96;
Δ = b2-4ac
Δ = 02-4·(-2)·96
Δ = 768
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:

$\sqrt{\Delta}=\sqrt{768}=\sqrt{256*3}=\sqrt{256}*\sqrt{3}=16\sqrt{3}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{3}}{2*-2}=\frac{0-16\sqrt{3}}{-4}
=-\frac{16\sqrt{3}}{-4}
=-\frac{4\sqrt{3}}{-1}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{3}}{2*-2}=\frac{0+16\sqrt{3}}{-4}
=\frac{16\sqrt{3}}{-4}
=\frac{4\sqrt{3}}{-1}
$