Solve the following equation:

$$
\frac{2\left(x^{2}-3\right)}{5-x^{2}}=\sqrt{\frac{5 x+6}{x+2}}
$$
The equation is undefined for $x= \pm \sqrt{5}$ and $x=-2$, so these values must be excluded. Squaring both sides,

$$
\frac{4\left(x^{4}-6 x^{2}+9\right)}{x^{4}-10 x^{2}+25}=\frac{5 x+6}{x+2}
$$

Removing the fractions and rearranging,

$$
x^{5}-2 x^{4}-26 x^{3}-12 x^{2}+89 x+78=0
$$

By a known theorem, if this equation has a rational root, it can only be among the divisors of 78. After a few trials, we can determine that $x_{1}=-1, x_{2}=2, x_{3}=-3$ satisfy the equation.

Dividing the polynomial on the left by $(x+1)(x-2)(x+3)=x^{3}-2 x^{2}-5 x-6$, we get $\left(x^{2}-4 x-13\right)$, so our equation can be written as:

$$
(x+1)(x-2)(x+3)\left(x^{2}-4 x-13\right)=0
$$

The missing two roots of the fifth-degree equation are provided by the

$$
x^{2}-4 x-13=0
$$

quadratic equation.

From this,

$$
x_{4}=2+\sqrt{17} \quad \text { and } \quad x_{5}=2-\sqrt{17} \text {. }
$$

Substitution shows that only

$$
x_{2}=2 \quad \text { and } \quad x_{5}=2-\sqrt{17}
$$

satisfy our original equation.