Find all 3 number solutions for $x[(x-2)^2+1]=6$ 
Find all 3 number solutions for $x[(x-2)^2+1]=6$

I used trial and error method to find integer solutions for $x$, and found that 1 possible solution is $x=3$.
However, there are 2 other non-integer solutions and I do not know how to find them.
I appreciate any help, thanks.

Looking for integer solutions, the equation $x[(x-2)^2+1]=6$  is equivalent to
$$\begin{cases}x=2,\\(x-2)^2+1=3, \end{cases}\qquad\text{or}\qquad\begin{cases}x=3,\\(x-2)^2+1=2. \end{cases}$$
The second equation  in the first  system implies that $(x-2)^2\equiv -1\mod 3$. Unfortunately, the only squares mos. $3$ are $0$ and $1$, so this first system has no solution.
The second equation  in the second  system means $(x-2)^2=1$, i.e. $x-2=\pm 1\iff x=3\;\text{ or }\;x=1 $. Only $x=3$ is compatible with the first equation.
Therefore there is a single integer solution. For the other solutions, we can expand the l.h.s. to obtain the cubic equation, divisible by $x-3$:
$$x^3-4x^2+5x-6=0\iff (x-3)(x^2-x+2)=0$$
The quadratic equation $x^2-x+2=0$ has complex conjugate roots:
$$x=\frac{1\pm i\sqrt 7}2.$$
