Решите уравнение: (x+3)(x-1)-(3x+1)(x-7)=x(x+18).

\[- 2x^{2} + 22x + 4 - x^{2} - 18x = 0\]

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

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

\[= 16 + 48 = 64\]

\[x_{1} = \frac{- 4 + \sqrt{64}\ }{2 \cdot ( - 3)} = \frac{- 4 + 8}{- 6} =\]

\[= \frac{4}{- 6} = - \frac{2}{3}\]

\[x_{2} = \frac{- 4 - \sqrt{64}\ }{2 \cdot ( - 3)} = \frac{- 4 - 8}{- 6} =\]

\[= \frac{- 12}{- 6} = 2\]

\[Ответ:\ x = - \frac{2}{3};x = 2.\]