jmc

Since the polynomial has no constant term, we can immediately factor out an $x$ from every term $$x(6x^3+19x^2-51x+20),$$and our first root $x=0$. Let $g(x)=6x^3+19x^2-51x+20$. Then the remaining roots of our original polynomial are the roots of $g(x)$. By trying out simple values, we can see that $g(0)=20>0$ and $g(1)=6+19-51+20=-6<0$. Thus, there must be a root of $g(x)$ between $0$ and $1$. From the Rational Root Theorem, we know that if $g(p/q)=0$ then $p$ must divide $20$ and $q$ must divide $6$.

Checking rational numbers of the form $p/q$, where $p$ divides $20$ and $q$ divides $6$, and $p/q$ is between $0$ and $1$, we find that $$\begin{array}{rl}g\left(\frac{1}{2}\right)& =6\cdot \frac{1}{8}+19\cdot \frac{1}{4}-51\cdot \frac{1}{2}+20=0.\end{array}$$This means that $2x-1$ is a factor of $g(x).$ Dividing by $2x-1$ gives us $g(x)=(2x-1)(3x^2+11x-20)$.

The quadratic $3x^2+11x-20$ factors as $(3x-4)(x+5),$ so our last two roots are $4/3$ and $-5$.

Thus, the roots of $6x^4+19x^3-51x^2+20x$ are $\boxed{0,\frac{1}{2},\frac{4}{3},-5}$.