To find the set of values of \(p\) for which the equation \(f(x) = p\) has no real roots, we consider the equation:

\(2x^2 - 6x + 5 - p = 0\)

This is a quadratic equation in the form \(ax^2 + bx + c = 0\) where \(a = 2\), \(b = -6\), and \(c = 5 - p\).

The discriminant \(\Delta\) of a quadratic equation \(ax^2 + bx + c = 0\) is given by:

\(\Delta = b^2 - 4ac\)

For the quadratic equation to have no real roots, the discriminant must be less than zero:

\(\Delta < 0\)

Substituting the values of \(a\), \(b\), and \(c\) into the discriminant:

\(\Delta = (-6)^2 - 4 \times 2 \times (5 - p)\)

\(\Delta = 36 - 8(5 - p)\)

\(\Delta = 36 - 40 + 8p\)

\(\Delta = 8p - 4\)

Setting \(\Delta < 0\):

\(8p - 4 < 0\)

\(8p < 4\)

\(p < \frac{1}{2}\)

Thus, the set of values of \(p\) for which the equation \(f(x) = p\) has no real roots is \(p < \frac{1}{2}\).