2w^2+6=96

Solution for 2w^2+6=96 equation:

2w^2+6=96

We move all terms to the left:

2w^2+6-(96)=0

We add all the numbers together, and all the variables

2w^2-90=0

a = 2; b = 0; c = -90;
Δ = b2-4ac
Δ = 02-4·2·(-90)
Δ = 720
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:

$\sqrt{\Delta}=\sqrt{720}=\sqrt{144*5}=\sqrt{144}*\sqrt{5}=12\sqrt{5}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{5}}{2*2}=\frac{0-12\sqrt{5}}{4}
=-\frac{12\sqrt{5}}{4}
=-3\sqrt{5}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{5}}{2*2}=\frac{0+12\sqrt{5}}{4}
=\frac{12\sqrt{5}}{4}
=3\sqrt{5}
$