3х^4-16х^2-12=0через x^2=t

Решите задачу:

$3x^4-16x^2-12=0\\\\t=x^2\\\\3t^2-16t-12=0\\\\t=\frac{-(-16)\pm\sqrt{(-16)^2-4*3*(-12)} }{2*3} =[-\frac{2}{3};\;6]\\\\\begin{minipage}{0.2\textwidth}$x$^2=-\frac{2}{3}\\\\x\notin \mathbb R\end{minipage}\hfill\begin{minipage}{0.9\textwidth}$x$^2=6\\\\x=\sqrt{6} \;\;\;\;\;x=-\sqrt{6}\end{minipage}\\\\\\Otvet: x_1=\sqrt{6}\;;\;\;x_2=-\sqrt{6}.$

3х^4-16х^2-12=0через x^2=t​

3х^4-16х^2-12=0через x^2=t