Решите уравнение: (3y^2+y-24)/(9-y^2 )=-2.

\[\frac{3y^{2} + y - 24}{9 - y^{2}} = - 2\]

\[ОДЗ:\ \ \ 9 - y^{2} \neq 0\]

\[\ \ \ \Longrightarrow y \neq \pm 3\]

\[3y^{2} + y - 24 = - 2 \cdot \left( 9 - y^{2} \right)\]

\[3y^{2} + y - 24 + 2 \cdot \left( 9 - y^{2} \right) = 0\]

\[3y^{2} + y - 24 + 18 - 2y^{2} = 0\]

\[y^{2} + y - 6 = 0\]

\[D = b^{2} - 4ac =\]

\[= 1 - 4 \cdot 1 \cdot ( - 6) = 25\]

\[y_{1} = \frac{- 1 + 5}{2} = \frac{4}{2} = 2\]

\[y_{2} = \frac{- 1 - 5}{2} = - \frac{6}{2} =\]

\[= - 3\ \ (не\ подходит\ по\ ОДЗ)\]

\[Ответ:y = 2.\]