What is the relevance of the two values of time here? So, I am in school and I was solving some questions related to Kinematics, those related to the three equations of motion. While solving them, I constantly faced  problems in the same kind of questions.
In questions where displacement ($S$), initial velocity ($u$) and acceleration ($a$) is given and time needs to be determined.
By $$S = ut+\frac{1}{2}at^2$$ We can find the time, but while solving, we encounter a quadratic. Now, when the quadratic yields one positive and one negative value of $t$, it becomes quite easy to reject the negative value of $t$, because $t$ can never be negative.
But I encountered some problems wherein $t$ was positive in both cases. For instance, $2+\sqrt{2}$ and $2-\sqrt{2}$. However I do not really understand which value to reject because I do not understand itself; what the individual times signify. How to know which value of time will match the requirements of the question and give us the correct answer? Besides, what is the relevance of the negative $t$ which we get in some quadratics?
 A: The equation $S=ut+\frac12at^2$ describes the position $S$ of an object moving in one dimension under constant acceleration $a$ as a function of time $t$, with a position of $0$ and a velocity of $u$ at $t=0$.
If we set $S$ to be some value $s$ and then solve for $t$, you are just finding all times where the position of the object is $s$. If there are multiple solutions, then that just means there are two instants in time where the object is at that position. Just like how one solution means there is one instant in time where the object is at that position, and no solutions means the object is never at that position.
As an explicit example, let's say the object is initially moving to the right with speed $u=1\,\mathrm{m/s}$, and is undergoing an acceleration of $-2\,\mathrm{m/s^2}$ (that is, an acceleration of $2\,\mathrm{m/s^2}$ to the left).

*

*If we wanted to find the times the object is at position $s=0\,\mathrm m$, we would find two times: $t=0\,\mathrm s$, and $t=1\,\mathrm s$; the object starts at $s=0$ moving to the right, turns around, and then comes back to the starting point in $1\,\mathrm s$.

*If we wanted to find the times the object is at position $s=0.25\,\mathrm m$, we would find one time $t=0.5\,\mathrm s$. This is the position where the object turns around, and so it makes sense that we only have 1 solution here; it is only at this position at one instant in time.

*If we wanted to find the times the object is at position $s=10\,\mathrm m$, we would find no solution. Since the object turned around at $s=0.25\,\mathrm m$, it never makes it to $s=10\,\mathrm m$, so it makes sense that we get no solution.

Note that, in general, negative time solutions are not necessarily unphysical, since we are not required to "start time" at $t=0$, but typically it is the case that (by convention) the motion started at $t=0$, and so negative time solutions do not work for the system in question. Going back to the above example, if we wanted to find the times the object is at position $s=-2\,\mathrm m$, we would find two times: $t=-1\,\mathrm s$, and $t=2\,\mathrm s$. The latter solution is when the object passes $s=-2\,\mathrm m$ after turning around. The former solution describes a situation where the object was in motion before $t=0$. e.g. if we had "started watching" the object $1\,\mathrm s$ before it initially passed $s=0\,\mathrm m$, we would have seen it at $s=-2\,\mathrm m$. If the object had actually started its motion at $t=0\,\mathrm s$ (say, someone launched the object at this time), then the negative time solution would indeed be unphysical.
A: Single values of time in introductory Newtonian mechanics are usually not negative (unless you define the initial time to be negative, but that's not common in intro mechanics to the best of my knowledge).
So when you use that equation for the displacement of a particle as a function of time, $S(t)$, assuming constant acceleration $a$, and you obtain two values of time that are both positive, this means that you have to think about which one makes sense and which one doesn't, or it's possible that there are two solutions in time, for example if we're talking about a ball bouncing. You did not give any specific examples so I cannot say more than this.
A: A specific example of this general question was asked recently (Here: Intuitive explanation for the second solution. Projectile problem).
The general equation:
$$ x(t) = x_0 + \dot x_0 t + \frac 1 2 \ddot x_0^2$$
has no knowledge of boundary conditions, so when you solve it:
$$ x(t) = x_0 + \dot x_0 t + \frac 1 2 \ddot x_0^2 =A(x-t_+)(x-t_-)$$
you get 2 roots at $t_{\pm}$. These are the two times on the parabolic trajectory (lasting for all time) such that:
$$ x(t_{\pm})=0$$
You just have to imagine that trajectory. If you're shooting a cannon ball off a building and ask when it will it the ground, you have to imagine a trajectory coming from infinity (with a flat Earth) that passes through the surface, moves through the cannon, at which point it becomes "your problem".
That both can be greater than zero is not odd, it just depends on how you set the problem up.
Obviously:
$$t_{\pm} = \frac{-\dot x_0\pm\sqrt{\dot x_0^2-2x_0\ddot x_0}}{2x_0}$$
so it depends on things like your coordinate system (e.g., where is $x=0$?), and cannot be physical.
