2. [6 points] Solve the equation $\sqrt{x+3}-\sqrt{7-x}+4=2 \sqrt{21+4 x-x^{2}}$.
Answer: $6, \frac{4-3 \sqrt{11}}{2}$.

Solution. Let $\sqrt{x+3}-\sqrt{7-x}=t$. Squaring both sides of this equation, we get $(x+3)-2 \sqrt{(x+3)(7-x)}+(7-x)=t^{2}$, from which $2 \sqrt{21+4 x-x^{2}}=10-t^{2}$. 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.

$$
\begin{aligned}
& \sqrt{x+3}-\sqrt{7-x}=2 \Leftrightarrow \sqrt{x+3}=2+\sqrt{7-x} \Leftrightarrow\left\{\begin{array}{l}
x+3=4+7-x+4 \sqrt{7-x} \\
-3 \leqslant x \leqslant 7
\end{array}\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ 2 \sqrt { 7 - x } = x - 4 , } \\
{ - 3 \leqslant x \leqslant 7 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
28-4 x=x^{2}-8 x+16 \\
4 \leqslant x \leqslant 7
\end{array}\right.\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ x ^ { 2 } - 4 x - 1 2 = 0 , } \\
{ 4 \leqslant x \leqslant 7 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x=-2 \text { or } x=6, \\
4 \leqslant x \leqslant 7
\end{array} \Leftrightarrow x=6\right.\right. \\
& \sqrt{x+3}-\sqrt{7-x}=-3 \Leftrightarrow \sqrt{x+3}+3=\sqrt{7-x} \Leftrightarrow\left\{\begin{array}{l}
x+3+6 \sqrt{x+3}+9=7-x \\
-3 \leqslant x \leqslant 7
\end{array}\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ 6 \sqrt { x + 3 } = - 5 - 2 x , } \\
{ - 3 \leqslant x \leqslant 7 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
36 x+108=25+20 x+4 x^{2} \\
-3 \leqslant x \leqslant-2.5
\end{array}\right.\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ 4 x ^ { 2 } - 1 6 x - 8 3 = 0 , } \\
{ - 3 \leqslant x \leqslant - 2 . 5 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x=\frac{4 \pm 3 \sqrt{11}}{2}, \\
-4 \leqslant x \leqslant-2.5
\end{array} \Leftrightarrow x=\frac{4-3 \sqrt{11}}{2} .\right.\right.
\end{aligned}
$$

Thus, the equation has two roots: $x=\frac{4-3 \sqrt{11}}{2}$ and $x=6$.