\[3x^{2} - 4x + n = 0;\ \ x_{1} = - 2\]
\[3\left( x^{2} - \frac{4}{3}x + \frac{1}{3}n \right) = 0\]
\[x_{1} + x_{2} = \frac{4}{3}\]
\[x_{2} = - 2 + 1\frac{1}{3}\]
\[x_{2} = - \frac{2}{3}.\]
\[x_{1} \cdot x_{2} = \frac{1}{3}n\]
\[\frac{1}{3}n = - 2 \cdot \left( - \frac{2}{3} \right)\]
\[\frac{1}{3}n = \frac{4}{3}\]
\[n = \frac{4}{3}\ :\frac{1}{3} = \frac{4}{3} \cdot 3 = 4.\]
\[Ответ:x_{2} = - \frac{2}{3};n = 4.\]