2g^2-32=32

Solution for 2g^2-32=32 equation:

2g^2-32=32

We move all terms to the left:

2g^2-32-(32)=0

We add all the numbers together, and all the variables

2g^2-64=0

a = 2; b = 0; c = -64;
Δ = b2-4ac
Δ = 02-4·2·(-64)
Δ = 512
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{2}}{2*2}=\frac{0-16\sqrt{2}}{4}
=-\frac{16\sqrt{2}}{4}
=-4\sqrt{2}
$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{2}}{2*2}=\frac{0+16\sqrt{2}}{4}
=\frac{16\sqrt{2}}{4}
=4\sqrt{2}
$