Let's arrange the equation as:
$$\left(x^2-9x+22\right)^2-9\left(x^2-9x+22\right)-x+22=0$$
Expanding and collecting like terms yields:
$$x^4-18x^3+116x^2-316x+308=0$$
Now, let's assume we may factor as follows:
$$x^4-18x^3+116x^2-316x+308=\left(x^2+ax+b\right)\left(x^2+cx+d\right)$$
Expand the right side:
$$x^4-18x^3+116x^2-316x+308=x^4+(a+c)x^3+(ac+b+d)x^2+(ad+bc)x+bd$$
Equating coefficients, we obtain the system:
$$a+c=-18$$
$$ac+b+d=116$$
$$ad+bc=-316$$
$$bd=308$$
Solving this system, we obtain:
$$a=-10,\,b=22,\,c=-8,\,d=14$$
Hence:
$$x^4-18x^3+116x^2-316x+308=\left(x^2-10x+22\right)\left(x^2-8x+14\right)$$
Application of the quadratic formula yields:
$$x=4\pm\sqrt{2},\,5\pm\sqrt{3}$$
Now, that is a mundane way to solve such a fun equation. Let's look at the substitution:
$$u=x^2-9x+22$$
And the original then becomes:
$$u^2-9u+22=x$$
Adding the two equations, we obtain:
$$u^2-8u+22=x^2-8x+22$$
And this may be reduced to:
$$(u-x)(u+x-8)=0$$
Back-substituting for $u$, we then get:
$$(x^2-9x+22-x)(x^2-9x+22+x-8)=0$$
$$(x^2-10x+22)(x^2-8x+14)=0$$
And the roots follow as above:
$$x=4\pm\sqrt{2},\,5\pm\sqrt{3}$$ :D