Найдите корни уравнения: (x^2-4)/3+4x=3.

\[\frac{x^{2} - 4}{3} + 4x = 3\ \ \ \ | \cdot 3\]

\[x^{2} - 4 + 12x = 9\]

\[x² + 12x - 13 = 0\ \]

\[D = b^{2} - 4ac =\]

\[= 144 - 4 \cdot 1 \cdot ( - 13) =\]

\[= 144 + 52 = 196\]

\[x_{1} = \frac{- 12 + 14}{2} = \frac{2}{2} = 1\]

\[x_{2} = \frac{- 12 - 14}{2} = - \frac{26}{2} = - 13\]

\[Ответ:x_{1} = 1;\ \ x_{2} = - 13.\]