Solution for the roots of $x^4+x^2+1=0$

Question

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:

Answer 1

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}