-4w^2+96=0

Solution for -4w^2+96=0 equation:

-4w^2+96=0

a = -4; b = 0; c = +96;
Δ = b2-4ac
Δ = 02-4·(-4)·96
Δ = 1536
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:

$\sqrt{\Delta}=\sqrt{1536}=\sqrt{256*6}=\sqrt{256}*\sqrt{6}=16\sqrt{6}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{6}}{2*-4}=\frac{0-16\sqrt{6}}{-8}
=-\frac{16\sqrt{6}}{-8}
=-\frac{2\sqrt{6}}{-1}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{6}}{2*-4}=\frac{0+16\sqrt{6}}{-8}
=\frac{16\sqrt{6}}{-8}
=\frac{2\sqrt{6}}{-1}
$