To find the \(x\)-coordinates of the intersection points, set the equations equal: \(7\sqrt{x} = 6x + 2\).

Rearrange to form a quadratic equation: \(6x + 2 = 7\sqrt{x} \Rightarrow 6(\sqrt{x})^2 - 7\sqrt{x} + 2 = 0\).

Let \(t = \sqrt{x}\), then the equation becomes \(6t^2 - 7t + 2 = 0\).

Factor the quadratic: \((3t - 2)(2t - 1) = 0\).

Solving for \(t\), we get \(t = \frac{2}{3}\) or \(t = \frac{1}{2}\).

Since \(t = \sqrt{x}\), we have \(\sqrt{x} = \frac{2}{3}\) or \(\sqrt{x} = \frac{1}{2}\).

Thus, \(x = \left(\frac{2}{3}\right)^2 = \frac{4}{9}\) or \(x = \left(\frac{1}{2}\right)^2 = \frac{1}{4}\).