Given the function \(f(x) = x^2 - 6x + c\), we need \(f(x) > 2\) for all \(x\).

Rearrange the inequality:

\(x^2 - 6x + c > 2\)

\(x^2 - 6x + c - 2 > 0\)

Complete the square for \(x^2 - 6x\):

\((x-3)^2 - 9\)

Substitute back:

\((x-3)^2 - 9 + c - 2 > 0\)

\((x-3)^2 + c - 11 > 0\)

Since \((x-3)^2 \geq 0\) for all \(x\), we need:

\(c - 11 > 0\)

Thus, \(c > 11\).