\[b_{1} = x - 4;\ b_{2} = \sqrt{7x};b_{3} = x + 6;\]
\[\frac{b_{2}}{b_{1}} = \frac{b_{3}}{b_{2} -}\]
\[\frac{\sqrt{7x}}{x - 4} = \frac{x + 6}{\sqrt{7x}};\ \ x > 0;x \neq 4\]
\[\left( \sqrt{7x} \right)^{2} = (x - 4)(x + 6)\]
\[7x = x^{2} - 4x + 6x - 24\]
\[x^{2} - 5x - 24 = 0\]
\[x_{1} + x_{2} = 5;\ \ x_{1} \cdot x_{2} = - 24\]
\[x_{1} = 8;\ \ x_{2} = - 3 < 0.\]
\[Ответ:x = 8.\]