6. Given functions
$$
f(x)=2 x^{2}+2 x-4, g(x)=x^{2}-x+2 \text {, }
$$

set $A=\left\{x \left\lvert\, \frac{f(x)}{g(x)} \in \mathbf{Z}_{+}\right.\right\}$.
Then $A=$ $\qquad$
6. $\left\{2, \frac{-3+\sqrt{33}}{2}, \frac{-3-\sqrt{33}}{2}\right\}$.

Notice that,
$$
\frac{f(x)}{g(x)}=\frac{2 x^{2}+2 x-4}{x^{2}-x+2}=2+\frac{4(x-2)}{x^{2}-x+2} \text {. }
$$

Also, since $x^{2}-x+2>0, x^{2}-5 x+10>0$, we know
$$
\begin{array}{l}
\frac{4(x-2)}{x^{2}-x+2}<1 \Rightarrow \frac{f(x)}{g(x)} \leqslant 2 \\
\Rightarrow \frac{f(x)}{g(x)}=1 \text { or } 2 \\
\Rightarrow A=\left\{2, \frac{-3+\sqrt{33}}{2}, \frac{-3-\sqrt{33}}{2}\right\} .
\end{array}
$$