Understand the symmetry

0 is not a root. (Constant term is not 0)

Hence we can divide the expression by $x^4$ .

We get a very ugly and long expression. However, if we let $a=x+\frac {2}{x}$ and substitute it, we obtain

$2a^{4}-9a^{3}+4a^{2}+21a-18=0$

Note that after some trial and error, 1, 2, 3 satisfy the equation. Thus we obtain

$2 (a-1)(a-2)(a-3)(a+1.5)=0$

after long division and the possible values of $a$ are 1, 2, 3, -1.5.

Now by AM-GM, $|x+\frac {2}{x}| \geq 2\sqrt {2}$

Hence the only possible value of $a$ such that $x$ is real is 3. And $x+\frac {2}{x}=3$ has two roots 1, 2. Hence the sum is 1+2=3. Nice question!