\[= - 12p^{2} - 8p + 4 \leq 0.\]
\[- 4 \cdot \left( 3p^{2} + 2p - ,1 \right) \leq 0\]
\[3p^{2} + 2p - 1 \geq 0\]
\[D = 4 + 12 = 16\]
\[p_{1,2} = \frac{- 2 \pm 4}{6} = \frac{1}{3}; - 1.\]
\[Ответ:\]
\[при\ \ p \in ( - \infty; - 1\rbrack \cup \left\lbrack \frac{1}{3}; + \infty \right).\]