2. Let the set $A=\left\{x \mid a x^{2}+3 x-2 a=0\right\}$ (where $a$ is a non-zero real number), $B=\left\{x \mid 2 x^{2}-5 x-42 \leqslant 0\right\}$, if $A \cap B=A$, then the range of values for $a$ is $\qquad$ .
Solve:
$2 x^{2}-5 x-42=(2 x+7)(x-6) \leqslant 0 \Rightarrow B=\left[-\frac{7}{2}, 6\right]$.
$\Delta=9+8 a^{2}>0 \Rightarrow f(x)=a x^{2}+3 x-2 a=0$ has two distinct real roots,
When $a>0$, $\left\{\begin{array}{l}\frac{49}{4} a-\frac{21}{2}-2 a \geqslant 0, \\ 36 a+18-2 a \geqslant 0, \\ -\frac{7}{2}<-\frac{3}{2 a}<6\end{array} \Rightarrow a \geqslant \frac{42}{41}\right.$;
When $a<0$, $\left\{\begin{array}{l}\frac{49}{4} a-\frac{21}{2}-2 a \leqslant 0, \\ 36 a+18-2 a \leqslant 0, \\ -\frac{7}{2}<-\frac{3}{2 a}<6\end{array} \Rightarrow a \leqslant-\frac{9}{17}\right.$.
Therefore, the range of $a$ is $\left(-\infty,-\frac{9}{17}\right] \cup\left[\frac{42}{41},+\infty\right)$.