\[\boxed{\mathbf{1140}\mathbf{\text{.\ }}Еуроки\ - \ ДЗ\ без\ мороки}\]
\[12x^{2} - bx + 5 = 0\]
\[x_{1} = - \frac{1}{3},\ \ x_{1}x_{2} = \frac{5}{12},\ \ \]
\[x_{2} = \frac{5}{12}\ :x_{1} = \frac{5}{12}\ :\left( - \frac{1}{3} \right) = - \frac{5}{4}\]
\[x_{1} + x_{2} = \frac{b}{12}\]
\[\frac{1}{3} - \frac{5}{4} = \frac{b}{12}\]
\[- \frac{19}{12} = \frac{b}{12}\]
\[b = - 19\]
\[Ответ:b = - 19;\ \ x_{2} = - \frac{5}{4}.\]