Solution for the roots of $x^4+x^2+1=0$ Is this solution to find the roots of $x^4+x^2+1=0$ correct?
$x^4+x^2+1=0$ 
$x^4+2x^2+1-x^2=0$
$(x^2+1)^2-x^2=0$ 
$[(x^2+1)-x][(x^2+1)+x]=0$
$(x^2-x+1)(x^2+x+1)=0$
For this equation to be true, either $(x^2-x+1)=0$ or/and $(x^2+x+1)=0$.
Using the quadratic formula, I got
$x=\frac{1\pm{\sqrt{3}i}}{2}$ 
and
$x=\frac{-1\pm{\sqrt{3}i}}{2}$ 
Are these values of x under the set of complex numbers the roots of $x^4+x^2+1=0$?
To answer this, I tried to check my answers using a computing website. According to the website, here are the solutions:


Given equation
\begin{eqnarray}
x^4+x^2+1=0
\end{eqnarray}
Suppose $x^2=y$ then $y^2+y+1=0$
using  quadratic formula we have
\begin{eqnarray}
y=\frac{-1\pm \sqrt{-3}}{2}\\
\end{eqnarray}
\begin{eqnarray}
Thus~~~~~~~~ x^2=\frac{-1\pm \sqrt{-3}}{2}\\
\Rightarrow  ~~x=\pm\sqrt\frac{-1\pm \sqrt{-3}}{2}
\end{eqnarray}
