3. Let's call a number $x$ semi-integer if the number $2x$ is an integer. The semi-integer part of a number $x$ is defined as the greatest semi-integer number not exceeding $x$, and we will denote it as $] x[$. Solve the equation $x^{2} + 2 \cdot ] x[ = 6$. (20 points)
Solution: Consider two cases.

1) The number $x-$ is half-integer, then $] x\left[=x\right.$ and the original equation will take the form $x^{2}+2 x-6=0$. The roots of this equation are $x_{1,2}=-1 \pm \sqrt{7}$, but then the numbers $2 x_{1,2}$ are not integers, so there are no solutions.
2) There is an equality $x=\frac{n}{2}+r$, where $n \in \mathbb{Z}$ and $00$ and can only have a solution if the inequality

$$
\frac{n}{2}0
\end{array}\right.
\end{gathered}
$$

The first inequality of the system is satisfied when $-2-2 \sqrt{7}-3+4 \sqrt{2}$ or $n4(6-n)>(n+1)^{2} \\
\left\{\begin{array}{l}
n^{2}+4 n-24>0 \\
n^{2}+6 n-23-2+2 \sqrt{7}$ or $n<-2-2 \sqrt{7}$. Since $-9<-3-4 \sqrt{2}<-8$, $-8<-2-2 \sqrt{7}<-7$, and $-3+4 \sqrt{2}<-2+2 \sqrt{7}$, then $n=-8-$ is the only integer value satisfying the system. In this case, $x=\frac{n}{2}+r=-\sqrt{6-n}=-\sqrt{14}$.

Answer: $\sqrt{3},-\sqrt{14}$