Решите уравнение: (4x+1)(x-3)=12.

\[(4x + 1)(x - 3) = 12\]

\[4x^{2} - 12x + x - 3 - 12 = 0\]

\[4x^{2} - 11x - 15 = 0\]

\[D = ( - 11)^{2} - 4 \cdot 4 \cdot ( - 15) =\]

\[= 121 + 240 = 361\]

\[x_{1} = \frac{- ( - 11) + \sqrt{361}}{2 \cdot 4} =\]

\[= \frac{11 + 19}{8} = \frac{30}{8} = \frac{15}{4} = 3\frac{3}{4}\]

\[x_{2} = \frac{- ( - 11) - \sqrt{361}}{2 \cdot 4} =\]

\[= \frac{11 - 19}{8} = \frac{- 8}{8} = - 1\]

\[Ответ:\ \ x = 3\frac{3}{4};\ x = - 1.\]