Solve for $x$: $x^4+2x^3-x-1=0$ Solve for  $x$:

$$x^4+2x^3-x-1=0$$

My try ; 

$$x^4+2x^3-x-1=(ax^2+bx+c)(dx^2+px+q)=0$$
But I cannot go further!!!
$$x^4+2x^3-x-1=\left(x^2+x-\frac{1}{2}\right)^2-\frac{5}{4}$$
and the rest is smooth.

We can get it by the following way.
$$x^4+2x^3-x-1=(x^2+x+k)^2-((2k+1)x^2+(2k+1)x+k^2+1)$$
for all value of $k$ and for $k=-\frac{1}{2}$ we'll get  a difference of squares.
hint

Let $f (x)=x^4+2x^3-x-1$.
$f (0)=-1,f (1)=1,f (-2)=1$
there is at least one root in $(0,1) $ and 

one in $ (-2,0)$.

$$f''(x)=12x^2+12x=12x (1+x)\geq 0$$

there are only two real roots.