How is final velocity negative when acceleration is in positive direction? When a body starts from rest and starts accelerating in positive direction and moves certain positive distance, its final velocity is given by :
$$v^2 = u^2 + 2as$$
Solving it gives two values of $v$, one positive and one negative.
But how can a body accelerating in  positive direction and starting from rest have a negative direction of its velocity? Am I missing something?
 A: There are two solutions for
$$v^2 = v_i^2 + 2a\Delta s,$$
one positive and one negative.  You can give a meaningful interpretation to both solutions. For example, for $v_i = 0 \textrm{ m/s}$, $a = 0.5 \textrm{ m/s}^2$, and $\Delta s = 2.0 \textrm{ m}$, we get
$$v = \pm \sqrt{2} \textrm{ m/s}.$$
The positive solution is easy to interpret. It is the speed of the particle at some time after the starting point. The negative solution can be interpreted in a symmetric way: it is the speed of the particle some time before the starting point.
So just imagine a particle moving in the negative direction while slowing down, coming to rest, and then moving in the positive direction while speeding up, with constant acceleration the entire time.
(Don't get hung up on 'starting from rest'. 'Starting from' just means that your analysis of the particle starts with initial values of position, velocity, etc. It does not need to imply that the particle had those same values for all times prior to your analysis of it.)
A: Your reasoning is not logically valid.
Your claim is the folowing:

"The answer to the problem is a solution to this equation."
$$\implies$$
"Every solution to this equation is an answer to the problem."

I see this as analogous to the following.
Given $x = 5$, we could reason that,
$$x = 5\implies x^2 = 25 \implies x = \pm\sqrt{25}= \pm5$$
But clearly $x \ne-5$, so even though $x = -5$ is a solution the equation $x^2 = 25$, we are wrong to conclude that it is true, given all our information (namely that $x = 5$ !).
My point is that it is often the case that you can reason for the equations to imply that there is additional solutions but, as is the case in maths as well as physics, we often have to return to the original conditions and question whether a solution is valid (e.g. "$f(x)$ is defined for all $x\in \mathbb{R}$" but then we find an solution to the equation $f(x) = 0$ but $x$ is complex - so we discard it).
In your case, a  negative velocity would not cause the body to move a positive distance, so it is to be discarded.
A: Maths is the language of physics not the other way around!
After this being said, you are right that velocity should have negative value even though a is positive, but ask yourselves does this solution makes practical sense?
Obviously no, so we discard the negative value.
There will be many scenarios where there would solutions that do not make any physical sense, just like 'hjjhk' is a word made of alphabets but does not have any meaning in English.
