5. Given positive numbers $a, b, c$ satisfy
$$
a \leqslant b+c \leqslant 3 a, 3 b^{2} \leqslant a(a+c) \leqslant 5 b^{2} \text {. }
$$

Then the minimum value of $\frac{b-2 c}{a}$ is $\qquad$
5. $-\frac{18}{5}$.

Transform the given inequality into $1 \leqslant \frac{b}{a}+\frac{c}{a} \leqslant 3, 3\left(\frac{b}{a}\right)^{2} \leqslant 1+\frac{c}{a} \leqslant 5\left(\frac{b}{a}\right)^{2}$. Let $\frac{b}{a}=x, \frac{c}{a}=y$. Then
$$
\left\{\begin{array}{l}
1 \leqslant x+y \leqslant 3, \\
3 x^{2} \leqslant 1+y \leqslant 5 x^{2}, \\
x>0, y>0 .
\end{array}\right.
$$

The region represented by this system of inequalities is the shaded area in Figure 3.