How to solve `14/(4x^2-1) + 3/(4x^2+4x+1)=5/(4x^2-4x+1)` ?

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`14/(4x^2-1) + 3/(4x^2+4x+1)=5/(4x^2-4x+1)`

First, factor the denominators.

`14/((2x-1)(2x+1))+3/((2x+1)(2x+1)) = 5/((2x-1)(2x-1))`

`14/((2x-1)(2x+1))+3/(2x+1)^2=5/((2x-1)^2)`

Then, multiply both sides by the LCD which is `(2x+1)^2(2x-1)^2` .

`(2x+1)^2(2x-1)^2*[14/((2x-1)(2x+1)) + 3/(2x+1)^2]=5/(2x-1)^2*(2x+1)^2(2x-1)^2`

`14(2x-1)(2x+1)+3(2x-1)^2=5(2x+1)^2`

`14(4x^2-1)+3(4x^2-4x+1)=5(4x^2+4x+1)`

`56x^2-14+12x^2-12x+3=20x^2+20x+5`

`68x^2-12x-11=20x^2+20x+5`

And express the equation in a form `ax^2+bx+c=0` . To do so, subtract both sides by `(20x^2+20x+5)` .

`68x^2-12x-11-(20x^2+20x+5)=20x^2+20x+5-(20x^2+20x+5)`

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