9. If the equation $x^{2}+(m+2) x+m+5=0$ has only positive roots, then the range of real number $m$ is $\qquad$ .
9. $(-5,-4]$.

Let the roots of the equation be $x_{1}, x_{2}$, then $\left\{\begin{array}{l}\Delta=(m+2)^{2}-4(m+5) \geqslant 0 \\ x_{1}+x_{2}=-(m+2)>0 \\ x_{1} x_{2}=m+5>0,\end{array}\right.$ we get $-5<m \leqslant-4$.