Answer
$\pm 1;\pm 2;\pm 4;\pm 5;\pm 10;\pm 20;\pm \dfrac {1}{2};\pm \dfrac {5}{2};\pm \dfrac {1}{3};\pm \dfrac {2}{3}\pm \dfrac {4}{3}\pm \dfrac {5}{3}\pm \dfrac {20}{3}\pm \dfrac {1}{6}\pm \dfrac {5}{6}\pm \dfrac {10}{3}$
Work Step by Step
In a polinomial function like
$f\left( x\right) =a_{n}x^{n}+a_{n-1}x^{n-1}+\ldots a_{1}x+a_{0}$
If $p/q$, in lowest terms, is a rational zero of $f$, then p must be a factor of $a_0 $ and $q$ must be a factor of $a_n$.
Here
$f\left( x\right) =6x^{4}+2x^{3}-x^{2}+20\Rightarrow a_{n}=6;a_{0}=20 $
Factors of $a_0$ are $\pm 1,\pm 2,\pm 4,\pm5, \pm 10,\pm20$
Factors of $a_n$ are $\pm 1,\pm 2,\pm3,\pm6$
So the potential rational zeros are
$\pm 1;\pm 2;\pm 4;\pm 5;\pm 10;\pm 20;\pm \dfrac {1}{2};\pm \dfrac {5}{2};\pm \dfrac {1}{3};\pm \dfrac {2}{3}\pm \dfrac {4}{3}\pm \dfrac {5}{3}\pm \dfrac {20}{3}\pm \dfrac {1}{6}\pm \dfrac {5}{6}\pm \dfrac {10}{3}$