6.130. In the equation $x^{2}-2 x+c=0$, determine the value of $c$ for which its roots $x_{1}$ and $x_{2}$ satisfy the condition $7 x_{2}-4 x_{1}=47$.
Solution.

From the condition by Vieta's theorem we have $\left\{\begin{array}{l}x_{1}+x_{2}=2, \\ x_{1} \cdot x_{2}=c, \\ 7 x_{2}-4 x_{1}=47 .\end{array}\right.$ From here, $x_{2}=2-x_{1}$ and we obtain

$$
\left\{\begin{array} { l } 
{ x _ { 1 } ( 2 - x _ { 1 } ) = c } \\
{ 7 ( 2 - x _ { 1 } ) - 4 x _ { 1 } = 4 7 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x_{1}\left(2-x_{1}\right)=c \\
x_{1}=-3
\end{array}\right.\right.
$$

Thus, $c=-15$.

Answer: $c=-15$.