Решите уравнение: 12m^2+m-6=0.

\[12m² + m - 6 = 0\ \]

\[a = 12,\ \ b = 1,\ \ c = - 6\]

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

\[= 1^{2} - 4 \cdot 12 \cdot ( - 6) =\]

\[= 1 + 288 = 289\]

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

\[m_{1} = \frac{- 1 + \sqrt{289}}{2 \cdot 12} = \frac{- 1 + 17}{12} =\]

\[= \frac{16}{24} = \frac{2}{3}\]

\[m_{2} = \frac{- 1 - \sqrt{289}}{2 \cdot 12} = \frac{- 1 - 17}{12} =\]

\[= - \frac{18}{12} = - \frac{3}{4}\]

\[Ответ:m = \frac{2}{3};m = - \frac{3}{4}.\]