\[x_{1} = \frac{2}{3}:\]
\[6x^{2} + bx - 3 = 0\]
\[\left\{ \begin{matrix} x_{1} + x_{2} = - \frac{b}{6} \\ x_{1} \cdot x_{2} = - \frac{3}{6}\text{\ \ \ } \\ \end{matrix} \right.\ \]
\[\frac{2}{3} \cdot x_{2} = - \frac{1}{2}\]
\[\ x_{2} = - \frac{1}{2} \cdot \frac{3}{2}\text{\ \ }\]
\[x_{2} = - \frac{3}{4}.\]
\[\frac{2}{3} + \left( - \frac{3}{4} \right) = - \frac{b}{6}\]
\[\frac{8}{12} - \frac{9}{12} = - \frac{b}{6}\text{\ \ }\]
\[- \frac{1}{12} = - \frac{b}{6}\]
\[\frac{b}{6} = \frac{1}{12}\text{\ \ }\]
\[b = \frac{6}{12}\]
\[b = \frac{1}{2}.\]
\[Ответ:\ \ b = \frac{1}{2};\ \ x_{2} = - \frac{3}{4}.\]