Solve the following equation:

$$
\frac{6}{\sqrt{x-8}-9}+\frac{1}{\sqrt{x-8}-4}+\frac{7}{\sqrt{x-8}+4}+\frac{12}{\sqrt{x-8}+9}=0
$$
Let the value of $\sqrt{x-8}$ be denoted by $y$, and first combine the two outer and the two middle terms on the left side of the equation:

$$
\frac{6}{y-9}+\frac{12}{y+9}=\frac{6(3 y-9)}{y^{2}-81} ; \quad \frac{1}{y-4}+\frac{7}{y+4}=\frac{8 y-24}{y^{2}-16}
$$

The numerators of both fractions are divisible by $2(y-3)$, so we factor this out before combining them. This gives us the equation

$$
\frac{2(y-3)\left[9\left(y^{2}-16\right)+4\left(y^{2}-81\right)\right]}{\left(y^{2}-16\right)\left(y^{2}-81\right)}=0
$$

One root of this equation is $y_{0}=3$, and the other roots could be the roots of the quadratic factor in the numerator, provided they are not roots of the denominator. These are the values $y_{1,2}= \pm 6$, which are indeed roots of (2). However, among the roots $y_{0}, y_{1}, y_{2}$, only non-negative values can be considered for $\sqrt{x-8}$. Therefore, the equation has two roots, $x_{0}=17$ and $x_{1}=44$.