Решите уравнение: 3y^2-13y+4=0.

\[3y² - 13y + 4 = 0\ \ \]

\[a = 3,\ \ b = - 13,\ \ c = 4\]

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

\[= ( - 13)^{2} - 4 \cdot 3 \cdot 4 =\]

\[= 169 - 48 = 121\]

\[y_{1,2} = \frac{- b \pm \sqrt{D}}{2a}\]

\[y_{1} = \frac{- ( - 13) + \sqrt{121}}{2 \cdot 3} =\]

\[= \frac{13 + 11}{6} = \frac{24}{6} = 4\]

\[y_{2} = \frac{- ( - 13) - \sqrt{121}}{2 \cdot 3} =\]

\[= \frac{13 - 11}{6} = \frac{2}{6} = \frac{1}{3}\]

\[Ответ:y = 4;y = \frac{1}{3}.\]