Example 2. Solve the inequality

$$
\frac{\sqrt{6+x-x^{2}}}{2 x+5} \geqslant \frac{\sqrt{6+x-x^{2}}}{x+4}
$$
Solution. The domain of definition of the original inequality is determined by the system

$$
\left\{\begin{array}{l}
6+x-x^{2} \geqslant 0 \\
2 x+5 \neq 0 \\
x+4 \neq 0
\end{array}\right.
$$

from which we find: $-2 \leqslant x \leqslant 3$.

For the values $x=-2$ and $x=3$, the inequality (2) is satisfied; therefore, these values are its solutions.

Let $-20$, $2 x+5>0,6+x-x^{2}>0$. Therefore, on the interval $(-2 ; 3)$, the original inequality is equivalent to the inequality

$$
x+4 \geqslant 2 x+5
$$

from which we find $x \leqslant-1$. Of these values of $x$, the interval $-2<x<3$ includes all numbers in the interval $-2<x \leqslant-1$.

Therefore, the solutions of the inequality (2) are all numbers from the interval $-2 \leqslant x \leqslant-1$ and $x=3$.