Answer: \(k\lt-15\) or \(k\gt-3\).

For the quadratic to have two distinct real roots, its discriminant must be positive.

Here

\(a=1,\qquad b=k+9,\qquad c=9.\)

So

\((k+9)^2-4(1)(9)\gt0.\)

This gives

\((k+9)^2-36\gt0.\)

Factorise as a difference of two squares:

\((k+9-6)(k+9+6)\gt0.\)

Therefore

\((k+3)(k+15)\gt0.\)

The product is positive outside the two critical values, so

\(k\lt-15\quad\text{or}\quad k\gt-3.\)