6. Let set $A=\left\{x \mid x^{2}-[x]=2\right\}, B=\{x \mid-2<x<2\}$, where the symbol $[x]$ denotes the greatest integer less than or equal to $x$, then $A \bigcap B=$ $\qquad$
Solve:
$[x]=x^{2}-2 \leqslant x \Rightarrow-1 \leqslant x \leqslant 2$, when $[x]=-1$, $x=-1$; when $[x]=0$, it does not meet the condition; when $[x]=1$, $x=\sqrt{3}$; when $[x]=2$, $x=2$, thus $A=\{-1, \sqrt{3}, 2\}$. Therefore, $A \cap B=\{-1, \sqrt{3}\}$.