2. [6 points] Solve the equation $\sqrt{x+1}-\sqrt{4-x}+3=2 \sqrt{4+3 x-x^{2}}$.
Answer: $3, \frac{3-2 \sqrt{6}}{2}$.

Solution. Let $\sqrt{x+1}-\sqrt{4-x}=t$. Squaring both sides of this equation, we get $(x+1)-2 \sqrt{(x+1)(4-x)}+(4-x)=t^{2}$, from which $2 \sqrt{4+3 x-x^{2}}=5-t^{2}$. 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.

$$
\begin{aligned}
& \sqrt{x+1}-\sqrt{4-x}=1 \Leftrightarrow \sqrt{x+1}=1+\sqrt{4-x} \Leftrightarrow\left\{\begin{array}{l}
x+1=1+4-x+2 \sqrt{4-x} \\
-1 \leqslant x \leqslant 4
\end{array}\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ \sqrt { 4 - x } = x - 2 , } \\
{ - 1 \leqslant x \leqslant 4 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
4-x=x^{2}-4 x+4 \\
2 \leqslant x \leqslant 4
\end{array}\right.\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ x ^ { 2 } - 3 x = 0 , } \\
{ 2 \leqslant x \leqslant 4 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x=0 \text { or } x=3, \\
2 \leqslant x \leqslant 4
\end{array} \quad \Leftrightarrow x=3 ;\right.\right. \\
& \sqrt{x+1}-\sqrt{4-x}=-2 \Leftrightarrow \sqrt{x+1}+2=\sqrt{4-x} \Leftrightarrow\left\{\begin{array}{l}
x+1+4 \sqrt{x+1}+4=4-x \\
-1 \leqslant x \leqslant 4
\end{array}\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ 4 \sqrt { x + 1 } = - 1 - 2 x , } \\
{ - 1 \leqslant x \leqslant 4 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
16 x+16=1+4 x+4 x^{2}, \\
-1 \leqslant x \leqslant-0.5
\end{array}\right.\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ 4 x ^ { 2 } - 1 2 x - 1 5 = 0 , } \\
{ - 1 \leqslant x \leqslant - 0 . 5 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x=\frac{3 \pm 2 \sqrt{6}}{2}, \\
-1 \leqslant x \leqslant-0.5
\end{array} \quad \Leftrightarrow x=\frac{3-2 \sqrt{6}}{2} .\right.\right.
\end{aligned}
$$

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