\[\frac{b^{2} - b - 6}{9b + 18} = \frac{(b - 3)(b + 2)}{9(b + 2)} =\]
\[= \frac{b - 3}{9}\]
\[b^{2} - b - 6 = 0\]
\[b_{1} + b_{2} = 1;\ \ \ b_{1} \cdot b_{2} = - 6\]
\[b_{1} = 3;\ \ \ b_{2} = - 2.\]