Assume $F$ is a field, and $f\in F[x]$ is polynomial.
To see that $f$ has a root in some extension of $F$, without loss of generality we can assume $f$ is irreducible. Indeed any polynomial $f$ is the product of some irreducible polynomials. So any root of any of the multipliers is also a root of $f$.
Now since $f$ is irreducible, the factor $K=F[x]/(f)$ is a field. We can see that the restriction of the natural homomorphism $\pi|_F:F\to K$ is injective, hence $F<K$. Now if $\bar x=x+(f)$ obviously $f(\bar x)=0$.
Is there any sensible way to identify $x+(f)$ with some element in the algebraic closure $\bar F$ of $F$? Is $K$ an algebraic extension? Is there a canonical injection $j:K\to\bar F$? If $F$ is a number field, is there any sensible way to identify $x+(f)$ with some complex number?
There is no canonical injection $K\to\overline F$.
Consider $F=\mathbb Q$ and $f=X^3-2$. The roots of $f$ in $\mathbb C$ are $\omega^k \sqrt[3]{2}$, where $\omega$ is a primitive cubic root of unit. Algebraically, these three roots are indistinguishable. You might be tempted to single the real one out, but there is no real reason to (no pun intended :-).
For a simpler example, $i$ and $-i$ are algebraically indistinguishable, as being the roots of $X^2+1$, but tradition has favored the one with positive imaginary part, possibly because positive rotations are taken counterclockwise by convention.