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

`(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))`