$$
\begin{array}{l}
\text { 6. Given the quadratic equation in } x \text { is } \\
x^{2}+a x+(m+1)(m+2)=0
\end{array}
$$

For any real number $a$, the equation has real roots. Then the range of real number $m$ is . $\qquad$
6. $-2 \leqslant m \leqslant-1$.

From the given condition, we know $\Delta=a^{2}-4(m+1)(m+2) \geqslant 0$, which means
$$
a^{2} \geqslant 4(m+1)(m+2)
$$

holds for all real numbers $a$.
Obviously, the minimum value of $a^{2}$ is 0.
Therefore, the above condition is equivalent to
$$
(m+1)(m+2) \leqslant 0 \Rightarrow-2 \leqslant m \leqslant-1 .
$$