10. Solve the equation $\sqrt{x^{2}+6 x+11}+\sqrt{x^{2}-6 x+11}=10$.
10. The original equation is transformed into $\sqrt{(x+3)^{2}+2}+\sqrt{(x-3)^{2}+2}=10$. By treating the constant "2" as a variable, let $2=y^{2}$, then $\sqrt{(x+3)^{2}+y^{2}}+\sqrt{(x-3)^{2}+y^{2}}=10$.
According to the definition of an ellipse, this equation represents an ellipse with foci at $F_{1}(-3,0), F_{2}(3,0)$ and a major axis length of 10, whose equation is $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$. Substituting $y^{2}=2$ back in, the solution to the original equation is $x= \pm \frac{5}{4} \sqrt{14}$.