To find the values of \(k\) for which the line \(y + kx = 12\) is a tangent to the curve \(y = 2x^2 - 8x + 14\), we first express the line equation in terms of \(y\):

\(y = -kx + 12\)

Set the expressions for \(y\) equal to each other:

\(2x^2 - 8x + 14 = -kx + 12\)

Rearrange to form a quadratic equation:

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

For the line to be tangent to the curve, the quadratic must have exactly one solution, which occurs when the discriminant is zero:

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

Here, \(a = 2\), \(b = k-8\), and \(c = 2\). Substitute these into the discriminant formula:

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

Simplify:

\((k-8)^2 = 16\)

Take the square root of both sides:

\(k-8 = 4 \quad \text{or} \quad k-8 = -4\)

Solve for \(k\):

\(k = 12 \quad \text{or} \quad k = 4\)