Решите уравнение: 3x^4+4x^3-14x^2+4x+3=0.

\[3x^{4} + 4x^{3} - 14x^{2} + 4x + 3 = 0\]

\[3x^{2} + 4x - 14 + \frac{4}{x} + \frac{3}{x^{2}} = 0\]

\[t = x + \frac{1}{x}\]

\[3t^{2} - 20 + 4t = 0\]

\[3t^{2} + 4t - 20 = 0\]

\[D = 4^{2} - 4 \cdot 3 \cdot ( - 20) =\]

\[= 16 + 240 = 256\]

\[t_{1} = \frac{- 4 + \sqrt{256}}{2 \cdot 3} = \frac{- 4 + 16}{6} =\]

\[= \frac{12}{6} = 2\]

\[t_{2} = \frac{- 4 - \sqrt{256}}{2 \cdot 3} = \frac{- 4 - 16}{6} =\]

\[= \frac{- 20}{6} = - \frac{10}{3}\]

\[1)\ x + \frac{1}{x} = 2\ \ \ \ \ | \cdot x\]

\[x^{2} - 2x + 1 = 0\]

\[D = ( - 2)^{2} - 4 \cdot 1 \cdot 1 = 4 - 4 =\]

\[= 0\]

\[x = \frac{2 + 0}{2} = 1.\]

\[x + \frac{1}{x} = \frac{- 10}{3}\ \ \ \ \ \ \ \ | \cdot 3x\]

\[2)\ 3x^{2} + 10x + 3 = 0\]

\[D = 10^{2} - 4 \cdot 3 \cdot 3 =\]

\[= 100 - 36 = 64\]

\[x_{1} = \frac{- 10 + \sqrt{64}}{2 \cdot 3} = \frac{- 10 + 8}{6} =\]

\[= \frac{- 2}{6} = - \frac{1}{3}\]

\[x_{2} = \frac{- 10 - \sqrt{64}}{2 \cdot 3} = \frac{- 10 - 8}{6} =\]

\[= \frac{- 18}{6} = - 3\]

\[Ответ:\ 1;\ - \frac{1}{3};\ - 3.\]