\[\boxed{\text{996.\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]
\[x^{2} - 6bx + 9b^{2} - 16 = 0\]
\[a = 1,\ \ b = - 6b,\ \ \]
\[c = {9b}^{2} - 16\]
\[D = b^{2} - 4ac = ( - 6b)^{2} -\]
\[- 4 \cdot 1 \cdot \left( 9b^{2} - 16 \right) = 36b^{2} -\]
\[- 36b^{2} + 64 = 64\]
\[x_{1,2} = \frac{- b \pm \sqrt{D}}{2a} = \frac{6b \pm \sqrt{64}}{2} =\]
\[= \frac{6b \pm 8}{2} = \frac{2 \cdot (3b \pm 4)}{2} =\]
\[= 3b \pm 4\]
\[\left\{ \begin{matrix} 3b + 4 < 0 \\ 3b - 4 < 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\left\{ \begin{matrix} 3b < - 4 \\ 3b > 4 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} b < - \frac{4}{3} \\ b < \frac{4}{3}\text{\ \ \ \ } \\ \end{matrix} \right.\ ,\ \ \left( - \infty;\ - 1\frac{1}{3} \right)\]
\[Ответ:при\ b \in \left( - \infty;\ - 1\frac{1}{3} \right)\ \]
\[уравнение\ имеет\ 2\ \]
\[отрицательных\ корня.\]