1. The problem is to solve the inequality $|x^2+2| \le 11$. 2. Recall that the absolute value inequality $|A| \le B$ means $-B \le A \le B$. 3. Applying this to our problem, we get: $$-11 \le x^2 + 2 \le 11$$ 4. Subtract 2 from all parts: $$-11 - 2 \le x^2 \le 11 - 2$$ $$-13 \le x^2 \le 9$$ 5. Since $x^2$ is always non-negative, the left inequality $-13 \le x^2$ is always true. 6. The right inequality $x^2 \le 9$ means: $$-3 \le x \le 3$$ 7. Therefore, the solution set is all $x$ such that $-3 \le x \le 3$. Final answer: $$\boxed{-3 \le x \le 3}$$