12. For a real number $x$, it is defined that $[x]=n$ if and only if $n \leqslant x<n+1$ $\left(n \in \mathbf{N}_{+}\right)$. Then the solution set of the inequality
$$
4[x]^{2}-36[x]+45<0
$$

is . $\qquad$
$12.2 \leqslant x<8$.
From $4[x]^{2}-36[x]+45<0$, we get $\frac{3}{2}<[x]<\frac{15}{2}$. Therefore, $2 \leqslant[x] \leqslant 7$. Hence, $2 \leqslant x<8$.