jmc

If we graph $y=x^2-(k-3)x-k+6,$ then we obtain an upward-facing parabola. Thus, the inequality $$x^2-(k-3)x-k+6>0$$holds as long as the discriminant of the quadratic is negative.

This gives us $$(k-3{)}^2-4(-k+6)<0.$$This simplifies to $k^2-2k-15<0,$ which factors as $(k+3)(k-5)<0.$ Thus, $k\in \boxed{(-3,5)}.$