(3x-1)(3x+1)=(x+1)(9x-5)

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Solution for (3x-1)(3x+1)=(x+1)(9x-5) equation: 



(3x-1)(3x+1)=(x+1)(9x-5)

We move all terms to the left:

(3x-1)(3x+1)-((x+1)(9x-5))=0

We use the square of the difference formula

9x^2-((x+1)(9x-5))-1=0

We multiply parentheses ..

9x^2-((+9x^2-5x+9x-5))-1=0



We calculate terms in parentheses: -((+9x^2-5x+9x-5)), so:

(+9x^2-5x+9x-5)

We get rid of parentheses

9x^2-5x+9x-5

We add all the numbers together, and all the variables

9x^2+4x-5

Back to the equation:

-(9x^2+4x-5)



We get rid of parentheses

9x^2-9x^2-4x+5-1=0

We add all the numbers together, and all the variables

-4x+4=0

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

-4x=-4

x=-4/-4

x=1

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