## Task B-2.2.

Determine the absolute value of the difference of the solutions of the equation

$$
x(10 a-5 x+6)=(5 a+1)(a+1), a \in \mathbf{R}
$$
## First solution.

After rearranging, the equation becomes a quadratic equation

$$
5 x^{2}-2(5 a+3) x+5 a^{2}+6 a+1=0
$$

1 point

Let $A=x_{1}-x_{2}$. Then,

$$
A^{2}=\left(x_{1}-x_{2}\right)^{2}=x_{1}^{2}-2 x_{1} x_{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}
$$

Apply Viète's formulas to the solutions $x_{1}, x_{2}$.

$$
x_{1}+x_{2}=\frac{2(5 a+3)}{5}, \quad x_{1} x_{2}=\frac{5 a^{2}+6 a+1}{5}
$$

Then,

$$
A^{2}=\left(\frac{2(5 a+3)}{5}\right)^{2}-4 \frac{5 a^{2}+6 a+1}{5}=\frac{4\left(25 a^{2}+30 a+9\right)-20\left(5 a^{2}+6 a+1\right)}{25}
$$

which simplifies to $A^{2}=\frac{16}{25}$, so $\left|x_{1}-x_{2}\right|=\frac{4}{5}$.