First, rewrite the equation \(2x^2 - 10x + 8 = kx\) as a standard quadratic equation:

\(2x^2 - (10 + k)x + 8 = 0\).

For the quadratic equation \(ax^2 + bx + c = 0\) to have no real roots, the discriminant must be less than zero:

\(b^2 - 4ac < 0\).

Here, \(a = 2\), \(b = -(10 + k)\), and \(c = 8\).

Calculate the discriminant:

\((-(10 + k))^2 - 4 \times 2 \times 8 < 0\)

\((10 + k)^2 - 64 < 0\)

\(100 + 20k + k^2 - 64 < 0\)

\(k^2 + 20k + 36 < 0\).

To find the values of \(k\), solve the inequality:

Factor the quadratic:

\((k + 18)(k + 2) < 0\).

The critical points are \(k = -18\) and \(k = -2\).

Test intervals around these points to determine where the inequality holds:

- For \(k < -18\), both factors are negative, so the product is positive.

- For \(-18 < k < -2\), one factor is negative and the other is positive, so the product is negative.

- For \(k > -2\), both factors are positive, so the product is positive.

Thus, the solution is \(-18 < k < -2\).