\[x^{2} + bx - 29 = 0;\ \ \ x_{1} = - x_{2}\]
\[\left\{ \begin{matrix} x_{1} + x_{2} = - b \\ x_{1} \cdot x_{2} = - 29 \\ \end{matrix} \right.\ \]
\[1)\ - x_{2} \cdot x_{2} = - 29\ \ \]
\[x_{2}^{2} = 29\]
\[x_{2} = \pm \sqrt{29}.\]
\[2)\ x_{1} = \pm \sqrt{29}.\]
\[3)\ - x_{2} + x_{2} = - b\ \ \]
\[b = 0.\]
\[Ответ:\ x = \pm \sqrt{29};\ \ \ b = 0.\]