1. **State the problem:** We need to find the values of $x$ for which the expression $\sqrt{3x+36}$ is defined. 2. **Recall the rule for square roots:** The square root function $\sqrt{y}$ is defined only when the radicand $y$ is greater than or equal to zero, i.e., $y \geq 0$. 3. **Apply the rule to our radicand:** We require $$3x + 36 \geq 0$$ 4. **Solve the inequality:** $$3x + 36 \geq 0$$ $$3x \geq -36$$ $$\cancel{3}x \geq \cancel{3}(-12)$$ $$x \geq -12$$ 5. **Interpretation:** The square root $\sqrt{3x+36}$ is defined for all real numbers $x$ such that $x$ is greater than or equal to $-12$. **Final answer:** $$\boxed{x \geq -12}$$