Решите уравнение: 21y^2-2y-3=0.

\[21y² - 2y - 3 = 0\]

\[D = ( - 2)^{2} - 4 \cdot 21 \cdot ( - 3) =\]

\[= 4 + 252 = 256\]

\[y_{1} = \frac{2 + \sqrt{256}}{2 \cdot 21} = \frac{2 + 16}{42} =\]

\[= \frac{18}{42} = \frac{3}{7}\]

\[y_{2} = \frac{2 - \sqrt{256}}{2 \cdot 21} = \frac{2 - 16}{42} =\]

\[= - \frac{14}{42} = - \frac{1}{3}\]

\[Ответ:\ x = \frac{3}{7};\ x = - \frac{1}{3}.\]