15. The solution set of the inequality
$$
-2<\sqrt{x^{2}-2 x+4}-\sqrt{x^{2}-10 x+28}<2
$$

is $\qquad$ .
15. $\{x \mid 3-\sqrt{2}<x<3+\sqrt{2}\}$.

The original inequality is
$$
\left|\sqrt{(x-1)^{2}+3}-\sqrt{(x-5)^{2}+3}\right|<2 \text {. }
$$

Let $3=y^{2}$, the inequality can be transformed into
$$
\left|\sqrt{(x-1)^{2}+y^{2}}-\sqrt{(x-5)^{2}+y^{2}}\right|<2 \text {. }
$$

By the definition of a hyperbola, the points satisfying the above condition lie in the region between the two branches of the hyperbola $(x-3)^{2}-\frac{y^{2}}{3}=1$. Therefore, the original inequality is equivalent to the system of inequalities $\left\{\begin{array}{l}(x-3)^{2}-\frac{y^{2}}{3} \leqslant 1, \\ y^{2}=3\end{array}\right.$. Hence, the solution set of the original inequality is $\{x \mid 3-\sqrt{2}<x<3+\sqrt{2}\}$.