jmc

In order for the quadratic $k{x}^2-3kx+4k+7=0$ to have real roots, its discriminant must be nonnegative. This gives us the inquality $$(-3k{)}^2-4(k)(4k+7)\ge 0.$$This expands as $-7k^2-28k\ge 0.$ This is equivalent to $k^2+4k\le 0,$ which factors as $k(k+4)\le 0.$ The solution to this inequality is $-4\le k\le 0.$ However, if $k=0,$ then the given equation is not quadratic, so the set of $k$ which works is $\boxed{[-4,0)}.$