jmc

First, $m(x)=\sqrt{x+5}+\sqrt{20-x}$ is always nonnegative.

Note that $$\begin{array}{rl}[m(x){]}^2& =x+5+2\sqrt{x+5}\cdot \sqrt{20-x}+20-x\\ & =25+2\sqrt{(x+5)(20-x)}\\ & =25+2\sqrt{-x^2+15x+100}\\ & =25+2\sqrt{\frac{625}{4}-{\left(x-\frac{15}{2}\right)}^2}.\end{array}$$Looking at the formula $$[m(x){]}^2=25+2\sqrt{(x+5)(20-x)},$$the square root $\sqrt{(x+5)(20-x)}$ is always nonnegative, so $[m(x){]}^2$ is at least 25, which means $m(x)$ is at least 5 (since $m(x)$ is always nonnegative). Furthermore, $m(-5)=\sqrt{0}+\sqrt{25}=5,$ so the minimum value of $m(x)$ is 5.

Looking at the formula $$[m(x){]}^2=25+2\sqrt{\frac{625}{4}-{\left(x-\frac{15}{2}\right)}^2},$$the expression under the square root attains its maximum when $x=\frac{15}{2}.$ At this value, $${\left[m\left(\frac{15}{2}\right)\right]}^2=25+2\sqrt{\frac{625}{4}}=50,$$so $m\left(\frac{15}{2}\right)=\sqrt{50}=5\sqrt{2}.$

Therefore, the range of the function is $\boxed{[5,5\sqrt{2}]}.$