2. Let $x_{1}$ and $x_{2}$ be the solutions of the quadratic equation

$$
6 x^{2}-5 x+1=0
$$

Without solving the given equation, form the quadratic equation in $y$ whose solutions are the numbers

$$
y_{1}=\frac{x_{1}+1}{x_{1}-1} \text { and } y_{2}=\frac{x_{2}+1}{x_{2}-1} .
$$
Solution. From equation (1) according to Vieta's formulas, we have $x_{1}+x_{2}=\frac{5}{6}$, $x_{1} x_{2}=\frac{1}{6}$. Let $y^{2}-p x+q=0$ be the desired equation; then again according to Vieta's formulas, we will have

$$
\begin{aligned}
p & =y_{1}+y_{2}=\frac{x_{1}+1}{x_{1}-1}+\frac{x_{2}+1}{x_{2}-1}=\frac{\left(x_{1}+1\right)\left(x_{2}-1\right)+\left(x_{2}+1\right)\left(x_{1}-1\right)}{\left(x_{1}-1\right)\left(x_{2}-1\right)} \\
& =\frac{2 x_{1} x_{2}-2}{x_{1} x_{2}-\left(x_{1}+x_{2}\right)+1}=-5 \\
q & =y_{1} \cdot y_{2}=\frac{x_{1}+1}{x_{1}-1} \cdot \frac{x_{2}+1}{x_{2}-1}=\frac{x_{1} x_{2}+\left(x_{1}+x_{2}\right)+1}{x_{1} x_{2}-\left(x_{1}+x_{2}\right)+1}=6
\end{aligned}
$$

Therefore, the desired equation is $y^{2}+5 y+6=0$.