4. Find all solutions of the equation $x^{2}-12 \cdot[x]+20=0$, where $[x]-$ is the greatest integer not exceeding $x$.
$$
\begin{aligned}
& \text { Solution. Let } [x]=a. \text { Then } a \leq x < a+1 \text { and } x^{2}+20=12 a . \text { We have: } \\
& \left\{\begin{array}{l}
x^{2}+20=12 a \\
a \leq x < a+1
\end{array} \Rightarrow\left\{\begin{array}{l}
x^{2}+20=12 a \\
a \leq x < a+1
\end{array} \Rightarrow\right.\right. \\
& \Rightarrow\left\{\begin{array}{l}
x^{2}+20=12 a \\
a \leq x < a+1
\end{array} \Rightarrow\left\{\begin{array}{l}
x^{2}+20=12 a \\
a \leq x < a+1
\end{array} \Rightarrow\right.\right. \\
& \Rightarrow\left\{\begin{array}{l}
(a-10)(a-2) \leq 0 \\
(a-3)(a-7)>0 ;
\end{array} \Rightarrow\right.\right. \\
& \Rightarrow\left\{\begin{array}{l}
a \in[2 ; 10], \\
a \in(-\infty ; 3) \cup(7 ;+\infty) ;
\end{array} \Rightarrow a \in\{2 ; 8 ; 9 ; 10\}\right.
\end{aligned}
$$

Since $x^{2}+20=12 a$, then $x$ takes the values $2 ; 2 \sqrt{19} ; 2 \sqrt{22} ; 10$.