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.
 A: $$x(x^2-4x+5)=6$$
$$x^3-4x^2+5x-6=0$$
$$(x-3)(x^2-x+2)=0$$
You just have to verify the discriminant of $x^2-x+2$ is negative and conclude that there is no other real root.
If you are interested to find the other roots, you might like to use the quadratic formula to find the remaining roots.
A: By educated trial & error:
If you assume that the exercise has an easy solution, an integer is likely. $6$ factors as $2\cdot3$ and as the second factor is a perfect square plus one, this rules out $3$. Then $x=3$ is a bingo !
Now shifting the unknown with $x:=z+3$, we obtain
$$z^3+5z^2+8z=0$$ or $$z\left(\left(z+\frac52\right)^2+\frac74\right)=0,$$ the resolution of which is easy.
A: 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:
$$x=\frac{1\pm i\sqrt 7}2.$$
A: When you look at the factors in the equation $ \ x · [ \ (x-2)^2 + 1 \ ] \ = \ 6 \ \ , $ it is possible to see that you will be not only out of integer solutions, but also out of real-number solutions.
Since we must have  $ \ (x-2)^2 \ + \ 1 \ \ge \ 1 \ \  $ for real values of $ \ x \ , $ then this being a positive number forces $ \ x \ $ as a factor to be positive only.  The possible remaining products of integer factors after $ \ 3 · 2 \ \ , $ which you found, are $ \ 1 · 6 \ , \ 2 · 3 \ , \ $ and $ \ 6 · 1 \ \ , $ but we can quickly check that $ \ (x-2)^2 \ + \ 1 \   $ does not produce the correct second factor for $ \ x = 1 , 2 , 6 \ \ . $
As $ \ x \ $ cannot equal zero, we can re-arrange the equation as $ \  (x-2)^2 + 1  \ = \ \frac{6}{x} \ \  $  and interpret it as the specification for the real-number $ \ x-$ coordinate(s) of intersection points between two curves.  The function $ \ y = \frac{6}{x} \ $ is a "rectangular" hyperbola with branches in the first and third quadrants; as we have established that $ \ x > 0 \ \ , $ we will only be concerned with the "positive branch".  This has the coordinate axes as its vertical and horizontal asymptotes.  The function $ \ y = (x-2)^2 + 1 \ $ is an "upward-opening" parabola with its vertex at $ \ (2 \ , \ 1 ) \ $ and its $ \ y-$ intercept at $ \ (0 \ , \ 5) \ \ . $  Since the hyperbola passes through $ \ (1 \ , \ 6) \ , \ (2 \ , \ 3 ) \ , \ (3 \ , \ 2) \ $ and $ \ (6 \ , \ 1 ) \ \ , $ it is already "above" the parabola at $ \ x = 2 \ $ and "below" it at $ \ x = 4 \ $ (with $ \ y-$coordinate $ \ y = \frac32 \ $ versus $ \ y = 5 \ $ for the parabola).  So there is only the one intersection point $ \ (3 \ , \ 2) \ $ which you've determined; a graph will readily confirm this.
Hence, $ \ x = 3 \ $ is the unique real-number solution to the equation in question.  (The two complex-number solutions are given in other answers.)
