1/2g^2-17=15

Solution for 1/2g^2-17=15 equation:

1/2g^2-17=15

We move all terms to the left:

1/2g^2-17-(15)=0


Domain of the equation: 2g^2!=0

g^2!=0/2

g^2!=√0

g!=0

g∈R


We add all the numbers together, and all the variables

1/2g^2-32=0

We multiply all the terms by the denominator


-32*2g^2+1=0

Wy multiply elements

-64g^2+1=0

a = -64; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-64)·1
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16}{2*-64}=\frac{-16}{-128}
=1/8
$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16}{2*-64}=\frac{16}{-128}
=-1/8
$