Answer
$x\in\left\{-2,-3\right\}$
Work Step by Step
We try to find rational zeros of $f(x)=x^{4}+4x^{3}+2x^{2}-x+6$
Possible rational roots: $\displaystyle \frac{p}{q}=\frac{\pm 1,\pm 2,\pm 3,\pm 6}{\pm 1}$
Testing with synthetic division, ... we try $x+2$
$\left.\begin{array}{l}
-2 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
1&4 & 2 & -1 &6 \\\hline
&-2 &-4 & 4 & -6 \\\hline
1&2 & -2 &3 & |\ \ 0 \end{array}$
$f(x)=(x+2)(x^{3}+2x^{2}-2x+3)$
Testing with synthetic division, ... we try $x+3$
$\left.\begin{array}{l}
-3 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
1 & 2 & -2 &3 \\\hline
&-3 & 3 & -3 \\\hline
1& -1 & 1 & |\ \ 0 \end{array}$
$f(x)=(x+2)(x+3)(x^{2}-x+1)$
The discriminant, $b^{2}-4ac=1-4=-3$
for the equation $ x^{2}-x+1=0\quad$ is negative,
so no further real zeros, as the equation does not have real solutions.
$f(x)=(x+2)(x+3)(x^{2}-x+1)=0$
when $x\in\left\{-2,-3\right\}$