1. The problem asks us to find the restrictions on the variable $x$ so that the expression $\sqrt{3-2x}$ is defined. 2. Recall that 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. Here, the radicand is $3-2x$. So, we set up the inequality: $$3 - 2x \geq 0$$ 4. Solve the inequality for $x$: $$3 - 2x \geq 0$$ Subtract 3 from both sides: $$-2x \geq -3$$ Divide both sides by $-2$, remembering to reverse the inequality sign because we are dividing by a negative number: $$\cancel{-2}x \leq \cancel{-3} \div -2$$ $$x \leq \frac{3}{2}$$ 5. Therefore, the expression $\sqrt{3-2x}$ is defined for all $x$ such that: $$x \leq \frac{3}{2}$$ This means $x$ can be any real number less than or equal to $1.5$. Final answer: $x \leq \frac{3}{2}$