5. If the equation with respect to $x$
$$
4^{x}+(a+3) 2^{x}+5=0
$$

has at least one real root in the interval $[1,2]$, then the range of the real number $a$ is $\qquad$ .
5. $-\frac{33}{4} \leqslant a \leqslant-3-2 \sqrt{5}$.

From $4^{x}+(a+3) 2^{x}+5=0$, we get $a+3=-\frac{4^{x}+5}{2^{x}}$. Let $t=2^{x}$. Since $x \in[1,2]$, hence $t \in[2,4]$, and $a+3=-\left(t+\frac{5}{t}\right)$.

The function $f(t)=t+\frac{5}{t}$ is decreasing on $[2, \sqrt{5}]$ and increasing on $[\sqrt{5}, 4]$; $f(2)=\frac{9}{2}, f(\sqrt{5})=2 \sqrt{5}, f(4)=$ $\frac{21}{4}$, so the range of $f(t)$ on $[2,4]$ is $\left[2 \sqrt{5}, \frac{21}{4}\right]$. Therefore, $-\frac{21}{4}$ $\leqslant a+3 \leqslant-2 \sqrt{5}$.
Thus, the range of the real number $a$ is $-\frac{33}{4} \leqslant a \leqslant-3-2 \sqrt{5}$.