# Solve for x. x^2+2x+3/2x-3 - 2x-3/x^2+2x+3 = 35/6Show complete solution and explain the answer

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`(x^2+2x+3)/(2x-3)-(2x-3)/(x^2+2x+3)=35/6`

Multiply by LCD = `(2x-3)(x^2+2x+3)`

`(x^2+2x+3)^2-(2x-3)^2=(35/6)(2x-3)(x^2+2x+3)`

Using difference of squares we get

`((x^2+2x+3)+(2x-3))((x^2+2x+3)-(2x-3))=35/6(2x-3)(x^2+2x+3)`

`6(x^2+4x)(x^2+6)=35(2x^3+4x^2+6x-3x^2-6x-9)`

Factor

`6x(x+4)(x^2+6)=35(2x^3+x^2-9)`

Multiply out

`6x(x^3+6x+4x^2+24)=70x^3+35x^2-315`

`6x^4+24x^3+36x^2+144x=70x^3+35x^2-315`

Puting in standard form

`6x^4-46x^3+x^2+144x+315=0`

We can graph to find possible roots: of 3 and 7

We can divide to find if it is a root

We can divide `6x^4-46x^3+x^2+144x+315` by (x-3) to see if it is a root.

3)6 -46 1 144 315

18 -84 -249 -315

6 -28 -83 -105 0

`(x-3)(6x^3-28x^2-83x-105)=0`

Now we can divide `6x^3-28x^2-83x-105` by (x-7) to find if we have another root.

7)6 -28 -83 -105

42 98 105

6 14 15 0

So `(x-3)(x-7)(6x^2+14x+15)=0`

We can use the quadratic formula to find the roots of `(6x^2+14x+15)` to find the full solution

x = 3, x = 7 and two imaginary roots `(-7/6 +- i(sqrt(41)/6))`