[tex]6x^4-5x^2+1=0\\
\text{Notam:} x^2=t\ \textgreater \ 0\\
6t^2-5t+1=0\\
\Delta=25-4\cdot6= 1\Rightarrow \sqrt{\Delta}=1\\
t_1=\dfrac{5+1}{12}=\dfrac{1}{2}\\
t_2=\dfrac{5-1}{12}=\dfrac{1}{3}\\
x^2=\dfrac{1}{2}\Rightarrow x\in \left\{\pm \dfrac{\sqrt{2}}{2}\right\}\\
x^2=\dfrac{1}{3}\Rightarrow x\in \left\{\pm \dfrac{\sqrt{3}}{3}\right\}\\
S:x\in \left\{ \pm \dfrac{\sqrt{3}}{3};\pm \dfrac{\sqrt{2}}{2}\right\}[/tex]
