Answer: \(-1\lt k\lt 3\).

For the quadratic to have no real roots, its discriminant must be negative.

Here \(a=4\), \(b=-4k\), and \(c=2k+3\). Therefore

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

So

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

This gives

\(16k^2-32k-48\lt 0.\)

Divide by \(16\):

\(k^2-2k-3\lt 0.\)

Factorise:

\((k-3)(k+1)\lt 0.\)

The expression is negative between the roots, so

\(\boxed{-1\lt k\lt 3}.\)