Решить уравнение: (x+1)(x+4)(x+7)(x+10)=63.

\[(x + 1)(x + 4)(x + 7)(x + 10) = 63\]

\[(x + 1)(x + 10)(x + 4)(x + 7) = 63\]

\[\left( x^{2} + x + 10x + 10 \right)\left( x^{2} + 4x + 7x + 28 \right) = 63\]

\[\left( x^{2} + 11x + 10 \right)\left( x^{2} + 11x + 28 \right) = 63\]

\[x^{2} + 11x = t:\]

\[(t + 10)(t + 28) = 63\]

\[t^{2} + 10t + 28t + 280 - 63 = 0\]

\[t^{2} + 38t + 217 = 0\]

\[D_{1} = 361 - 217 = 144\]

\[t_{1} = - 19 + 12 = - 7;\]

\[t_{2} = - 19 - 12 = - 31.\]

\[1)\ x^{2} + 11x = - 7\]

\[x^{2} + 11x + 7 = 0\]

\[D = 121 - 28 = 93\]

\[x_{1,2} = \frac{- 11 \pm \sqrt{93}}{2}.\]

\[2)\ x^{2} + 11x = - 31\]

\[x^{2} + 11x + 31 = 0\]

\[D = 121 - 124 < 0\]

\[нет\ корней.\]

\[Ответ:x = \frac{- 11 \pm \sqrt{93}}{2}.\]