A quartic with no real roots

SOS decomposition.
$$x^4+2x^3+3x^2+2x+1=(x^2+x)^2+2x^2+2x+1?$$
Let me check: $(x^2+x)^2=x^4+2x^3+x^2$. Then we need to add $2x^2+2x+1$. So
$$P(x)=(x^2+x)^2+(x+1)^2+x^2?$$
Check: $(x^2+x)^2=x^4+2x^3+x^2$, $(x+1)^2=x^2+2x+1$, $x^2=x^2$. Sum: $x^4+2x^3+3x^2+2x+1$. Correct.

Conclusion. $P(x)=(x^2+x)^2+(x+1)^2+x^2\ge0$ for all $x$, and $P(x)=0$ iff $x^2+x=0$, $x+1=0$, $x=0$, which is impossible ($x=0$ and $x=-1$). So $P(x)>0$ for all $x$, no real roots.