Example 4 Inequality
$$
-2<\sqrt{x^{2}-2 x+4}-\sqrt{x^{2}-10 x+28}<2
$$

The solution set is $\qquad$ .
(2006, National High School Mathematics League, Heilongjiang Province Preliminary Contest)
Solve The original inequality can be transformed into
$\left|\sqrt{x^{2}-2 x+4}-\sqrt{x^{2}-10 x+28}\right|<2$.
Let $y=\sqrt{3}$. Then the inequality becomes
$\left|\sqrt{(x-1)^{2}+(y-0)^{2}}-\sqrt{(x-5)^{2}+(y-0)^{2}}\right|<2$
The above can be seen as the absolute value of the difference in distances from the point $(x, y)$ to the points $(1,0)$ and $(5,0)$ being less than 2.

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

Therefore, the solution set of the original inequality is
$$
\{x \mid 3-\sqrt{2}<x<3+\sqrt{2}\} \text {. }
$$