6. Let set $A=\left\{x \mid x^{2}-[x]=2\right\}$, $B=\{x|| x \mid<2\}$,
where, $[x]$ denotes the greatest integer less than or equal to $x$.
Then $A \cap B=$ $\qquad$
$Two \mathbf{6} . A \cap B=\{-1, \sqrt{3}\}$.
Since $|x|<2$, therefore, the value of $[x]$ can be -2, $-1,0,1$.
When $[x]=-2$, $x^{2}=0$, no solution;
When $[x]=-1$, $x^{2}=1 \Rightarrow x=-1$;
When $[x]=0$, $x^{2}=2$ no solution;
When $[x]=1$, $x^{2}=3 \Rightarrow x=\sqrt{3}$.
Therefore, $x=-1$ or $\sqrt{3}$.