Because $2$ is co-prime to $11$, we find that any solution $x$ must be co-prime to $11$.

By Fermat's little theorem, we have $x^{10}\equiv 1\pmod{11}$. It follows that $$x^{22} + x^{11}\equiv x^2 + x  \equiv x^2 + 12x + 2\equiv (x+6)^2 - 36\equiv 2\pmod{11}$$

This is equivalent to $(x+6)^2\equiv 5\pmod{11}$. Testing $(x+6)=0$, $\pm 1$, $\pm 2$, $\pm 3$, $\pm 4$, and $\pm 5$ finds $(x + 6) = \pm 4$ are solutions. Hence, the solutions to the given equation are $x\equiv 1\pmod{11}$ or $x\equiv 9\pmod{11}$.