No Roots of Negatives

Over the real numbers, a square root accepts zero and positive numbers only. So for a rule like $f\left(x\right) = \sqrt{x-5}$, the inside must satisfy $x-5 \ge 0$. Solving gives $x \ge 5$: the domain is 5 and every number above it.

This time the answer is an inequality, not a single excluded number. A whole stretch of inputs is refused, everything below 5.

Watch the sign when x has a negative coefficient. For $g\left(x\right) = \sqrt{10-2x}$, solve $10-2x \ge 0$: subtracting 10 gives $-2x \ge -10$, and dividing by -2 flips the sign, so $x \le 5$. Check with 0: the inside is 10, which is fine.