366. Solve the equation:

$$
\left(\frac{x+6}{x-1}\right)^{2}+3 \frac{x-1}{x+6}=\frac{515}{8}
$$
$\triangle$ The substitution $\frac{x+6}{x-1}=y$ suggests itself. We obtain:

$$
y^{2}+\frac{3}{y}=\frac{515}{8}, \quad 8 y^{3}-515 y+24=0
$$

To find the integer roots of the last equation, we need to check the divisors of the constant term 24. We can notice that if the equation has an integer root, then it is even. Among the even divisors of 24, 8 fits (check it!). After dividing the left-hand side by $y-8$, we get:

$$
(y-8)\left(8 y^{2}+64 y-3\right)=0
$$

From this, we have:

$$
y_{1}=8, \quad y_{2,3}=\frac{-16 \pm \sqrt{262}}{4}
$$

Now $x$ can be easily found.

Answer: $2, \frac{-211 \pm 14 \sqrt{262}}{69}$.