Example 5. Solve the equation $\sqrt{x^{2}+6 x+36}$
$$
=\sqrt{x^{2}-14 x+76}+8 \text{. }
$$
Substitute $y^{2}$ for 27, the original equation becomes
$$
\sqrt{(x+3)^{2}+y^{2}}-\sqrt{(x-7)^{2}+y^{2}}=8 \text {. }
$$

This indicates that point $p(x, y)$ lies on the right branch of a hyperbola with foci at $F_{1}(-3,0)$ and $F_{2}(7,0)$:
$$
\frac{(x-2)^{2}}{16}-\frac{y^{2}}{9}=1
$$

Let $y^{2}=27$, we get $x=10$, which is the root of the original equation.