Решите уравнение: (корень из x-7)(24x^2-14x-3)=0.

\[\left( \sqrt{x} - 7 \right)\left( 24x^{2} - 14x - 3 \right) = 0\]

\[\sqrt{x} = 7\]

\[x = 49.\]

\[D = ( - 14)^{2} - 4 \cdot 24 \cdot ( - 3) =\]

\[= 196 + 288 = 484\]

\[x_{1} = \frac{14 + \sqrt{484}}{2 \cdot 24} = \frac{14 + 22}{48} =\]

\[= \frac{36}{48} = \frac{3}{4}\]

\[x_{2} = \frac{14 - \sqrt{484}}{2 \cdot 24} = \frac{14 - 22}{48} =\]

\[= - \frac{8}{48} = - \frac{1}{6}\]

\[Ответ:x = 49;\ x = \frac{3}{4};\ x = - \frac{1}{6}.\]