(1-3x/1+3x)+(1+3x/3x-1)=12/1-9x^2

Solution for (1-3x/1+3x)+(1+3x/3x-1)=12/1-9x^2 equation:

(1-3x/1+3x)+(1+3x/3x-1)=12/1-9x^2

We move all terms to the left:

(1-3x/1+3x)+(1+3x/3x-1)-(12/1-9x^2)=0


Domain of the equation: 1-9x^2)!=0

We move all terms containing x to the left, all other terms to the right

-9x^2)!=-1

x!=-1/1

x!=-1

x∈R



Domain of the equation: 1+3x)!=0

We move all terms containing x to the left, all other terms to the right

3x)!=-1

x!=-1/1

x!=-1

x∈R



Domain of the equation: 3x-1)!=0

x∈R


We add all the numbers together, and all the variables

-(12/1-9x^2)+(3x-3x/1+1)+(+3x/3x)=0

We get rid of parentheses

9x^2+3x-3x/1+3x/3x+1-12/1=0

Fractions to decimals

9x^2-3x/1+3x+1+1-12/1=0

We multiply all the terms by the denominator


9x^2*1-3x+3x*1-12+1*1+1*1=0

We add all the numbers together, and all the variables

9x^2*1-3x+3x*1-10=0

Wy multiply elements

9x^2-3x+3x-10=0

We add all the numbers together, and all the variables

9x^2-10=0

a = 9; b = 0; c = -10;
Δ = b2-4ac
Δ = 02-4·9·(-10)
Δ = 360
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:

$\sqrt{\Delta}=\sqrt{360}=\sqrt{36*10}=\sqrt{36}*\sqrt{10}=6\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{10}}{2*9}=\frac{0-6\sqrt{10}}{18}
=-\frac{6\sqrt{10}}{18}
=-\frac{\sqrt{10}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{10}}{2*9}=\frac{0+6\sqrt{10}}{18}
=\frac{6\sqrt{10}}{18}
=\frac{\sqrt{10}}{3}
$