How exactly is $i=\sqrt{-1}$ related to $\mathbb{C}$ being a closed algebraic field? There are many known proofs of why $\mathbb{C}$ (field of complex numbers) is algebraically closed (for example Cauchy's proof )
However:

how does introducing the solution to the equation $x^2 + 1 = 0$ (imaginary
  number $i$) makes this possible or is intimatelly related to it?

Thanx
 A: Adding a square root of any other negative number would work just as well. For $a>0$, if $x$ satisfies $x^2+a=0$, then $y=x/\sqrt a$ satisfies $y^2+1=0$. This is only trivially different. 
You could also do things like add on a solution to $x^4+1=0$, and then it would square to a solution to $x^2+1=0$, etc. (Edit: To clarify, as Servaes said, any polynomial that doesn't have a root in $\mathbb R$ would work. )I don't know if you would consider this sort of thing "trivially different" or if it answers your question. The thing is that adding on a root to $x^4+1=0$ (call it $j$) basically gives you the same complex numbers (the algebraic closure is essentially unique). In this particular example, every number of the form $a+bj$ can essentially (well, you can choose some signs) be written as $\left(a+\frac{b}{\sqrt 2}\right)+\frac{b}{\sqrt 2}i$ and every number of the form $a+bi$ can essentially be written as $(a-b)+\left(\sqrt 2 b\right)i$.
I don't know whether this is what you're looking for or not, but the Artin-Schreier Theorem basically states that if you only need to add on finitely many things to make your field algebraically closed, then adding on a zero of $x^2+1$ will do the job. A nice write-up is here.
A: Other replies have answered the meat of this question.  I just wanted to point out that if you start talking about matrix equations, instead of just polynomials, you end up with more fun things that extend the complex numbers.  
http://en.wikipedia.org/wiki/Quaternion
http://www.math.mcgill.ca/bsmith/Revisedpaper.pdf
