Answer: \(k=10\pm\sqrt{48}\).

The key point is to differentiate first, then substitute the required value or solve the resulting equation.

At an intersection of the line and the curve,

\(4x^2+kx+k-2=2x+1.\)

Rearrange:

\(4x^2+(k-2)x+k-3=0.\)

For the line to be a tangent, this quadratic must have equal roots. Therefore its discriminant is zero:

\((k-2)^2-4(4)(k-3)=0.\)

Expand and simplify:

\(k^2-4k+4-16k+48=0.\)

So

\(k^2-20k+52=0.\)

Using the quadratic formula,

\(k=\frac{20\pm\sqrt{(-20)^2-4(1)(52)}}{2}.\)

Thus

\(k=\frac{20\pm\sqrt{192}}{2} =10\pm\sqrt{48}.\)