-(2x^2-96)/(3x^2)=0

Solution for -(2x^2-96)/(3x^2)=0 equation:

-(2x^2-96)/(3x^2)=0


Domain of the equation: 3x^2!=0

x^2!=0/3

x^2!=√0

x!=0

x∈R


We multiply all the terms by the denominator


-(2x^2-96)=0

We get rid of parentheses

-2x^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:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{768}=\sqrt{256*3}=\sqrt{256}*\sqrt{3}=16\sqrt{3}$
$x_{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}
$
$x_{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}
$