Task B-1.7. Determine all natural numbers $n$ for which the fraction $\frac{6 n^{2}-10 n-12}{3 n-5}$ is a natural number.
Solution. Let's write the given fraction in a more convenient form:

$$
\begin{aligned}
\frac{6 n^{2}-10 n-12}{3 n-5} & =\frac{2 n(3 n-5)-12}{3 n-5} \\
& =2 n-\frac{12}{3 n-5}
\end{aligned}
$$

For the obtained expression to be a natural number, it must hold that

$$
\begin{aligned}
(3 n-5) & \in\{1,-1,2,-2,3,-3,4,-4,6,-6,12,-12\} \\
3 n & \in\{6,4,7,3,8,2,9,1,11,-1,17,-7\} \\
n & \in\{1,2,3\}
\end{aligned}
$$

Since $\frac{6 n^{2}-10 n-12}{3 n-5} \in \mathbb{N}$, it must be $2 n>\frac{12}{3 n-5}$

so the solutions are $n=1$ or $n=3$.

(1 point)