$\sqrt9$ has two solutions? This question might seem to be a little ridiculous, but... 
$\sqrt9=3$ right? it's obvious, but
$\sqrt9=-3$ what has happened here? 
We raise $-3$ to second power and we have 9, so nothing really wrong happened here?
What kind of knowlege am I missing? 
Even wolfram says that :$3$ is principal root and $-3$ is real root.
Thanks for any answer. 
 A: If $z$ is a nonzero complex number, there are precisely two complex numbers $w_1$ and $w_2$ such that $w_1^2 = w_2^2 = z$.  We have $w_2 = -w_1$ always.  (Indeed, one can show that the equation $x^2-z = 0$ can have at most two solutions in the complex numbers, and if $w_1^2 = z$, then $(-w_1)^2 = w_1^2 = z$.)  These are called the square roots of z.  (You can show this, for instance, by considering complex numbers in "polar form" $z = r e^{i \theta}$.  It is also a very special case of the Fundamental Theorem of Algebra.)
If $z$ happens to be a positive real number, then both $w_1$ and $w_2$ will be real, and thus one of them is positive and one of them is negative.  (Actually, the fact that every positive real number has a positive square root is used in the polar form decomposition above.  This basic fact can be proved using the Intermediate Value Theorem.)  It is a useful standard convention to denote by $\sqrt{z}$ the positive number, but you should remember that $-\sqrt{z}$ also counts as a square root of $z$, 
i.e., a number whose square is equal to $z$.  
Finally, note that for an arbitrary complex number $z$ the symbol $\sqrt{z}$ does not have a standard definition.  Here the selection of one out of the two square roots cannot be done as nicely as in the positive real case.  For instance, suppose you decide that you always want $\sqrt{z}$ to have an argument in $[0,\pi)$.  You can do so, but something bad happens just as you prepare to close up the unit circle near the border of the fourth and first quadrants.  If your complex number $z$ has argument slightly smaller than $2 \pi$, then the square root you have chosen has argument slightly smaller than $\pi$, but as soon as you hit $z= 1$ the argument of the square root jumps discontinuously to $0$.  In fact it is not possible to define a continuous function $f$ from the complex plane to itself so that $f(z)^2 = z$ for all $z \in \mathbb{C}$.  This phenomenon is an important one in both undergraduate complex analysis -- wherein it leads to branch cuts -- and in graduate complex analysis -- wherein it leads to Riemann surfaces.  So when someone -- or some software program -- speaks of the square root of a non-real complex number, you should learn to be a bit skeptical: something fishy may be going on.
A: You are missing the fact that, for $y>0$, $x^2=y\iff x=\pm \sqrt{y}$. 
So, there is two possible values for $x$.
A: Every number* has two square roots (except maybe zero, and even then you might claim that the roots are 0 and 0). 
However, we have made a special function called the square-root function, and it just spits out the "principle root"-- in the case of positive numbers, it is the positive root.
*If you insist that complex roots don't count, then just change this to every positive number. But this doesn't change the spirit of what's going on.
