Answer: \(k\lt -4\) or \(k\gt 4\).

At an intersection point, the two expressions for \(y\) are equal:

\(2x^2+k+4=(k+4)x.\)

Rearranging gives the quadratic equation

\(2x^2-(k+4)x+(k+4)=0.\)

For the curve and the line to intersect at two distinct points, this quadratic must have two distinct real roots. Therefore its discriminant must be positive:

\([-(k+4)]^2-4(2)(k+4)\gt 0.\)

So

\((k+4)^2-8(k+4)\gt 0.\)

Factorising,

\((k+4)(k-4)\gt 0.\)

This product is positive when both factors have the same sign, so

\(k\lt -4\quad\text{or}\quad k\gt 4.\)