2. [6 points] Solve the equation $\sqrt{x+4}-\sqrt{6-x}+4=2 \sqrt{24+2 x-x^{2}}$.
Answer: $5, \frac{2-3 \sqrt{11}}{2}$.

Solution. Let $\sqrt{x+4}-\sqrt{6-x}=t$. Squaring both sides of this equation, we get $(x+4)-2 \sqrt{(x+4)(6-x)}+(6-x)=t^{2}$, from which $2 \sqrt{24+2 x-x^{2}}=10-t^{2}$. The equation becomes $t+4=10-t^{2}$; hence $t^{2}+t-6=0$, i.e., $t=2$ or $t=-3$. We consider each case separately.

$$
\begin{aligned}
& \sqrt{x+4}-\sqrt{6-x}=2 \Leftrightarrow \sqrt{x+4}=2+\sqrt{6-x} \Leftrightarrow\left\{\begin{array}{l}
x+4=4+6-x+4 \sqrt{6-x} \\
-4 \leqslant x \leqslant 6
\end{array}\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ 2 \sqrt { 6 - x } = x - 3 , } \\
{ - 4 \leqslant x \leqslant 6 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
24-4 x=x^{2}-6 x+9, \\
3 \leqslant x \leqslant 6
\end{array}\right.\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ x ^ { 2 } - 2 x - 1 5 = 0 , } \\
{ 3 \leqslant x \leqslant 6 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x=-3 \text { or } x=5, \\
3 \leqslant x \leqslant 6
\end{array} \quad \Leftrightarrow x=5 ;\right.\right. \\
& \sqrt{x+4}-\sqrt{6-x}=-3 \Leftrightarrow \sqrt{x+4}+3=\sqrt{6-x} \Leftrightarrow\left\{\begin{array}{l}
x+4+6 \sqrt{x+4}+9=6-x, \\
-4 \leqslant x \leqslant 6
\end{array}\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ 6 \sqrt { x + 4 } = - 7 - 2 x , } \\
{ - 4 \leqslant x \leqslant 6 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
36 x+144=49+28 x+4 x^{2}, \\
-4 \leqslant x \leqslant-3.5
\end{array}\right.\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ 4 x ^ { 2 } - 8 x - 9 5 = 0 , } \\
{ - 4 \leqslant x \leqslant - 3 . 5 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x=\frac{2 \pm 3 \sqrt{11}}{2}, \\
-4 \leqslant x \leqslant-3.5
\end{array} \Leftrightarrow x=\frac{2-3 \sqrt{11}}{2} .\right.\right.
\end{aligned}
$$

Thus, the equation has two roots: $x=\frac{2-3 \sqrt{11}}{2}$ and $x=5$.