$\begin{equation*} \begin{cases} x_2 - x_1 = 4\frac{1}{3} &\\ x_1 + x_2 = -\frac{b}{3} &\\ x_1x_2 = \frac{10}{3} & \end{cases} \end{equation*}$
$\begin{equation*} \begin{cases} 2x_2 = \frac{13}{3} - \frac{b}{3} = \frac{13 - b}{3} &\\ 2x_1 = -\frac{b}{3} - \frac{13}{3} = -\frac{13 + b}{3} &\\ x_1x_2 = \frac{10}{3} & \end{cases} \end{equation*}$
$x_1x_2 = -\frac{13 + b}{6} * \frac{13 - b}{6} = \frac{169 - b^2}{36} = \frac{10}{3}$
$\frac{b^2 - 169}{10} = \frac{36}{3} = 12$
$b^2 - 169 = 120$
$b^2 - 169 = 120$
$b^2 = 289$
$b_{1,2} = ±17$
Ответ: ±17