The challenge here is to find a solution that doesn't use calculus, i.e., doesn't take the derivative. The key insight is that, as you translate a polynomial graph up or down, the number of roots changes only when you get to a translate with a double root. So what we're looking for is values of $k$ such that
for some root $r$ and coefficients $b$ and $c$.
Re-expanding the right hand side to
we see that
Thus $b=4+6r$ and $c=2br-3r^2-12=2(4+6r)r-3r^2-12=9r^2+8r-12$ and
so the number of roots changes when $k=cr^2=(9r^2+8r-12)r^2$ with $r=0$, $1$, and $-2$, i.e, when $k=0$, $5$, and $32$. By thinking about the general nature of quartics, we see that $3x^4+4x^3-12x^2+k$ has two roots when $k\lt0$, four roots (counting multiplicities) when $0\le k\le5$, two roots again when $5\lt k\le32$, and no roots when $k\gt32$. In particular, the answer to the question posed is the range $0\le k\le5$.
Comparing this to the answers that use the derivative, the take-home lesson may be that it's well worth learning calculus!