## Task B-2.1.

Determine the real parameter $m$ such that for the solutions $x_{1}, x_{2}$ of the equation

$$
5 x^{2}-10 m^{2} x-m x-3 x+6 m^{2}+m-4=0
$$

it holds that $\left(5 x_{1}-1\right)\left(5 x_{2}-1\right)=2018$.
## Solution.

Let's introduce the notations $a=5, b=-10 m^{2}-m-3, c=6 m^{2}+m-4$.

By Vieta's formulas, we have

$$
x_{1}+x_{2}=\frac{-b}{a}=\frac{10 m^{2}+m+3}{5}, \quad x_{1} x_{2}=\frac{c}{a}=\frac{6 m^{2}+m-4}{5}
$$

It is given that $\left(5 x_{1}-1\right)\left(5 x_{2}-1\right)=2018$, which we can rewrite in a more convenient form:

$$
25 x_{1} x_{2}-5\left(x_{1}+x_{2}\right)+1=2018
$$

Therefore, we have

$$
\begin{aligned}
& 5\left(6 m^{2}+m-4\right)-10 m^{2}-m-3=2017 \\
& \Longleftrightarrow 20 m^{2}+4 m-2040=0 \\
& \Longleftrightarrow 5 m^{2}+m-510=0
\end{aligned}
$$

Solving this quadratic equation, we get two possibilities:

$$
m_{1}=10, \quad m_{2}=-\frac{51}{5}
$$