To find the value of \(k\) for which the line \(y = 2x + k\) is a tangent to the curve \(y = 2x^2 - 6x + 5\), we equate the two equations:

\(2x^2 - 6x + 5 = 2x + k\)

Rearrange to form a quadratic equation:

\(2x^2 - 8x + 5 - k = 0\)

For the line to be tangent to the curve, the discriminant of this quadratic equation must be zero:

\(b^2 - 4ac = 0\)

Here, \(a = 2\), \(b = -8\), and \(c = 5 - k\).

Calculate the discriminant:

\((-8)^2 - 4(2)(5 - k) = 0\)

\(64 - 8(5 - k) = 0\)

\(64 - 40 + 8k = 0\)

\(24 + 8k = 0\)

\(8k = -24\)

\(k = -3\)

Thus, the value of \(k\) is \(-3\).