13. Find the range of the function $f(x)=2 x+3+\sqrt{-2 x^{2}+12 x-14}$.
13. Let $f(x)=m, y=-2 x-3+m$, then $y=\sqrt{-2 x^{2}+12 x-14} \geqslant 0$, rearranging gives $\frac{y^{2}}{4}+\frac{(x-3)^{2}}{2}=1(y \geqslant$
0 ). Since $y=-2 x+m-3$ is a set of parallel lines with a slope of -2, $y=\sqrt{-2 x^{2}+12-14}$ is the upper part of an ellipse with its major axis parallel to the $y$-axis above the $x$-axis.

From the graph, we know that when the line passes through the point $(3-\sqrt{2}, 0)$, $m$ takes its minimum value $9-2 \sqrt{2}$; when the line is tangent to the ellipse, $m$ is at its maximum. By $-2 x+m-3=\sqrt{-2 x^{2}+12-14}$ having equal roots, i.e., $6 x^{2}-4 m x+m^{2}-6 m+23=0$ having equal roots, thus $\Delta=m^{2}-18 m+69=0$, we get $m=9+2 \sqrt{3}$ (discard the smaller value), the range is $[9-2 \sqrt{2}, 9+2 \sqrt{3}]$.