Решите уравнение: 3x^4+16x^2-12=0.

\[3x^{4} + 16x² - 12 = 0\]

\[Пусть\ \ t = x^{2} \geq 0:\]

\[3t^{2} + 16t - 12 = 0\]

\[D = 256 + 144 = 400\]

\[t_{1} = \frac{- 16 + 20}{6} = \frac{4}{6} = \frac{2}{3};\ \ \]

\[t_{2} = \frac{- 16 - 20}{6} =\]

\[= - 6 < 0\ (не\ подходит).\]

\[Подставим:\ \]

\[x^{2} = \frac{2}{3}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\]

\[x = \pm \sqrt{\frac{2}{3}}\text{\ \ \ \ \ \ \ \ \ \ \ }\]

\[Ответ:\ x = \pm \sqrt{\frac{2}{3}}.\]