Answer: $2, \frac{1-2 \sqrt{6}}{2}$.

Solution. Let $\sqrt{x+2}-\sqrt{3-x}=t$. Squaring both sides of this equation, we get $(x+2)-2 \sqrt{(x+2)(3-x)}+(3-x)=t^{2}$, from which $2 \sqrt{6+x-x^{2}}=5-t^{2}$. The equation becomes $t+3=5-t^{2}$; hence $t^{2}+t-2=0$, i.e., $t=1$ or $t=-2$. We consider each case separately.

$$\begin{aligned} & \sqrt{x+2}-\sqrt{3-x}=1 \Leftrightarrow \sqrt{x+2}=1+\sqrt{3-x} \Leftrightarrow\left\{\begin{array}{l} x+2=1+3-x+2 \sqrt{3-x} \\ -2 \leqslant x \leqslant 3 \end{array}\right. \\ & \Leftrightarrow\left\{\begin{array} { l } { \sqrt { 3 - x } = x - 1 , } \\ { - 2 \leqslant x \leqslant 3 } \end{array} \Leftrightarrow \left\{\begin{array}{l} 3-x=x^{2}-2 x+1, \\ 1 \leqslant x \leqslant 3 \end{array}\right.\right. \\ & \Leftrightarrow\left\{\begin{array} { l } { x ^ { 2 } - x - 2 = 0 , } \\ { 1 \leqslant x \leqslant 3 } \end{array} \Leftrightarrow \left\{\begin{array}{l} x=-1 \text { or } x=2, \\ 1 \leqslant x \leqslant 3 \end{array} \quad \Leftrightarrow x=2\right.\right. \\ & \sqrt{x+2}-\sqrt{3-x}=-2 \Leftrightarrow \sqrt{x+2}+2=\sqrt{3-x} \Leftrightarrow\left\{\begin{array}{l} x+2+4 \sqrt{x+2}+4=3-x \\ -2 \leqslant x \leqslant 3 \end{array}\right. \\ & \Leftrightarrow\left\{\begin{array} { l } { 4 \sqrt { x + 2 } = - 3 - 2 x , } \\ { - 2 \leqslant x \leqslant 3 } \end{array} \Leftrightarrow \left\{\begin{array}{l} 16 x+32=9+12 x+4 x^{2}, \\ -2 \leqslant x \leqslant-1.5 \end{array}\right.\right. \\ & \Leftrightarrow\left\{\begin{array} { l } { 4 x ^ { 2 } - 4 x - 23 = 0 , } \\ { - 2 \leqslant x \leqslant - 1.5 } \end{array} \Leftrightarrow \left\{\begin{array}{l} x=\frac{1 \pm 2 \sqrt{6}}{2}, \\ -2 \leqslant x \leqslant-1.5 \end{array} \quad \Leftrightarrow x=\frac{1-2 \sqrt{6}}{2} .\right.\right. \end{aligned}$$

Thus, the equation has two roots: $x=\frac{1-2 \sqrt{6}}{2}$ and $x=2$.