(3x2+1)/3x2=1+1=2

Solution for (3x2+1)/3x2=1+1=2 equation:

(3x^2+1)/3x^2=1+1=2

We move all terms to the left:

(3x^2+1)/3x^2-(1+1)=0


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

x^2!=0/3

x^2!=√0

x!=0

x∈R


We add all the numbers together, and all the variables

(3x^2+1)/3x^2-2=0

We multiply all the terms by the denominator


(3x^2+1)-2*3x^2=0

Wy multiply elements

-6x^2+(3x^2+1)=0

We get rid of parentheses

-6x^2+3x^2+1=0

We add all the numbers together, and all the variables

-3x^2+1=0

a = -3; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-3)·1
Δ = 12
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:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{3}}{2*-3}=\frac{0-2\sqrt{3}}{-6}
=-\frac{2\sqrt{3}}{-6}
=-\frac{\sqrt{3}}{-3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{3}}{2*-3}=\frac{0+2\sqrt{3}}{-6}
=\frac{2\sqrt{3}}{-6}
=\frac{\sqrt{3}}{-3}
$