\[\frac{7b^{2} + 11b - 6}{9 - 49b^{2}} = - \frac{(7b - 3)(b + 2)}{(7b - 3)(7b + 3)} =\]
\[= - \frac{b + 3}{7b + 3}\]
\[7b^{2} + 11b - 6 = 7 \cdot \left( b - \frac{3}{7} \right)(b + 2) =\]
\[= (7b - 3)(b + 2)\]
\[D = 121 + 168 = 289\]
\[b_{1} = \frac{- 11 + 17}{14} = \frac{6}{14} = \frac{3}{7};\ \ \ \]
\[b_{2} = \frac{- 11 - 17}{14} = - \frac{28}{14} = - 2.\]