3. Solve the inequality:

$$
x^{2}+4 x \sqrt{x+6} \leqslant 5(x+6)
$$
3. The initial condition is $x+6 \geqslant 0$ (i.e., $x \geqslant -6$). Adding the expression $4(x+6)$ to both sides of the equation, we get $x^{2}+4 x \sqrt{x+6}+4(x+6) \leqslant 9(x+6)$, i.e., $(x+2 \sqrt{x+6})^{2} \leqslant 9(x+6)$, which simplifies to $|x+2 \sqrt{x+6}| \leqslant 3 \sqrt{x+6}$, i.e., $-3 \sqrt{x+6} \leqslant x+2 \sqrt{x+6} \leqslant 3 \sqrt{x+6}$.

First, let's solve the inequality $-3 \sqrt{x+6} \leqslant x+2 \sqrt{x+6}$, i.e., $-5 \sqrt{x+6} \leqslant x$. This inequality is always satisfied for $x \geqslant 0$, while for $x < 0$ after squaring, we get $x^{2} \leqslant x+6$, i.e., $x^{2}-x-6 \leqslant 0$; the function on the left side has zeros at points $\frac{1 \pm \sqrt{1+24}}{2}=\frac{1 \pm 5}{2}$, so for $x > 0$ it is non-negative on the interval $x \in (0,3]$. Therefore, the solution to the observed inequality is $x \in [-6,3]$.

All together, the solution to the inequality in the problem statement is:

$$
x \in [-5, \infty) \cap [-6,3] = [-5,3]
$$