Ответ:

\[\left\{ \begin{matrix} 1 > 0 \\ D < 0 \\ \end{matrix} \right.\ \]

\[= - 4a^{2} + 12a < 0\]
\[- 4a^{2} + 12a = 0\]
\[- 4a(a - 3) = 0\]
\[a = 0,\ \ a = 3\]

\[Ответ:( - \infty;0) \cup (3;\ + \infty).\]

\[\left\{ \begin{matrix} 1 > 0 \\ D < 0 \\ \end{matrix} \right.\ \]

\[= - 4a^{2} + 12a < 0\]
\[- 4a^{2} + 12a = 0\]
\[- 4a(a - 3) = 0\]
\[a = 0,\ \ a = 3\]

\[Ответ:( - \infty;0) \cup (3;\ + \infty).\]