`(1+6/(x^2))^2 -11(1+6/(x^2))+28 = 0`

This might look like very difficult, but we can make a simple substitution and make it easier.

Let `t = 1+6/(x^2)`

Then the above equation changes into,

t^2-11t+28 = 0

(t -7)(t-4) = 0

Now there are two solutions for t, they are t=7 or t=4.

Let's first solve for x using t =7,

`t =7 = 1+6/(x^2)`

This gives,

`7x^2=x^2+6`

`6x^2 = 6`

`x^2 = 1.`   -----> `x = +-sqrt(1) = +-1`

Therefore, when t =7, x = +1 or x =-1.

When t =4,

`t =4 = 1+6/(x^2)`

`4x^2=x^2+6`

`3x^2 = 6`

`x^2 = 2` ----------------> `x = +-sqrt(2)`

Therefore, when `t = 4` , `x = +sqrt(2) or -sqrt(2).`

The complete solutions for x are,

`x = +1` , `x=-1` , `x =+sqrt(2)` and `x =-sqrt(2)`

The solution set is,

`x = {-sqrt(2),-1,1,sqrt(2)}`