2. If $\sqrt{7 x^{2}+9 x+13}+\sqrt{7 x^{2}-5 x+13}=$ $7 x$, then $x=$ $\qquad$
2. $\frac{12}{7}$.

Obviously, $x>0$. From the condition, we have
$$
\frac{14 x}{\sqrt{7 x^{2}+9 x+13}-\sqrt{7 x^{2}-5 x+13}}=7 x \text {. }
$$

Then $\sqrt{7 x^{2}+9 x+13}-\sqrt{7 x^{2}-5 x+13}=2$.
Thus, $2 \sqrt{7 x^{2}+9 x+13}=7 x+2$.
Squaring and simplifying, we get $21 x^{2}-8 x-48=0$.
Therefore, the positive root is $x=\frac{12}{7}$.