7. Regulation: For $x \in \mathbf{R}$, if and only if $n \leqslant x < n+1\left(n \in \mathbf{Z}_{+}\right)$, then $[x]=n$. Then
$$
4[x]^{2}-36[x]+45 \leqslant 0
$$

the solution set is . $\qquad$
Two, $7.2 \leqslant x<8$.
Notice that, the required inequality is a quadratic inequality in terms of $[x]$.
$$
\begin{array}{l}
\text { By } 4[x]^{2}-36[x]+45 \leqslant 0 \\
\Rightarrow \frac{3}{2} \leqslant[x] \leqslant \frac{15}{2} \\
\Rightarrow 2 \leqslant[x] \leqslant 7 \\
\Rightarrow 2 \leqslant x<8 .
\end{array}
$$