Решите уравнение: 4x^4-21x^2+5=0.

\[4x^{4} - 21x^{2} + 5 = 0,\]

\[Пусть\ t = x^{2} \geq 0\]

\[4t^{2} - 21t + 5 = 0\]

\[D = 441 - 80 = 361 = 19^{2}\]

\[t_{1} = \frac{21 + 19}{8} = 5;\ \ \]

\[t_{2} = \frac{21 - 19}{8} = \frac{1}{4}.\]

\[Подставим:\ \]

\[x^{2} = 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x^{2} = \frac{1}{4}\]

\[x = \pm \sqrt{5}\ \ \ \ \ \ \ \ \ \ \ x = \pm \frac{1}{2}\]

\[Ответ:\ x = \pm \sqrt{5};\ x = \pm \frac{1}{2}.\]