Answer
$x\displaystyle \in\left\{\frac{1}{2},\ 2,\ 5\right\}$
Work Step by Step
We try to find rational zeros of $f(x)=2x^{4}-19x^{3}+57x^{2}-64x+20$
There are at most 4 real zeros.
Possible rational roots: $\displaystyle \frac{p}{q}=\frac{\pm 1,\pm 2,\pm 4,\pm 5,\pm 10,\pm 20}{\pm 1,\pm 2}$
Testing with synthetic division, ... we eventually try $x-2$
$\left.\begin{array}{l}
2 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
2 & -19 & 57 & -64 &20 \\\hline
& 4 & -30 & 54 & -20 \\\hline
2& -15 & 27 &-10 & |\ \ 0 \end{array}$
$f(x)=(x-2)(2x^{3}-15x^{2}+27x-10)$
Testing with synthetic division, ... again, try $x-2$
$\left.\begin{array}{l}
2 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
2 & -15 & 27 &-10 \\\hline
& 4 & -22 & +10 \\\hline
2& -11 &5 & |\ \ 0 \end{array}$
$f(x)=(x-2)^{2}(2x^{2}-11x+5)$
Testing with synthetic division, ... try $x-\displaystyle \frac{1}{2}$
$\left.\begin{array}{l}
1/2 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
2 & -11 &5 \\\hline
& 1 & -5 \\\hline
2&-10 & |\ \ 0 \end{array}$
$f(x)=(x-2)^{2}(x-\displaystyle \frac{1}{2})\cdot 2(x-5)=0 $
when $x\displaystyle \in\left\{\frac{1}{2},\ 2,\ 5\right\}$
