Simplifying: \(2|\sqrt{x}-3| + x - 6\sqrt{x} + 6 = 0\). Let \(t = \sqrt{x}\), so \(2|t-3| + t^2 - 6t + 6 = 0\). For \(t \geq 3\): \(2(t-3) + t^2 - 6t + 6 = 0\), giving \(t^2 - 4t = 0\), so \(t=0\) or \(t=4\). Only \(t=4\) satisfies \(t \geq 3\). For \(t < 3\): \(2(3-t) + t^2 - 6t + 6 = 0\), giving \(t^2 - 8t + 12 = 0\), so \(t=2\) or \(t=6\). Only \(t=2\) satisfies \(t<3\). Thus \(t \in \{2,4\}\), giving \(x \in \{4, 16\}\). S contains 2 elements.