In order to explain why you get two roots, and how we know which is the "right" one, let me consider the system of two carts with the spring attached to them at both ends (I have reason to believe that this is not what the question intended, but it is equivalent to what you are doing with the conservation laws). Let $M = m_1 + m_2$ be the total mass of the carts and $t = 0$ be the moment the string gets burned and so the spring starts to uncompress. Going straight to Newton's laws, we then have a pair of differential equations:
m_1\ddot x_1 &= k(x_2 - x_1)\\
m_2\ddot x_2 &= k(x_1 - x_2)
This can be easily decoupled by rewriting in terms of $X = (m_1/M)x_1 + (m_2/M)x_2$ (this is the position of center of mass) and $Y = x_2-x_1$ (the distance between the carts). The first one obeys $\ddot X = 0$, which means the center of mass keeps going right at constant velocity (equal to the initial $v$) and expresses the same thing as conservation of momentum. The second one obeys $\ddot Y + \omega^2 Y = 0$ where $\omega$ is a quantity that can be expressed in terms of the masses and the spring constant (it is the angular frequency of oscillations of the system of carts connected by the spring). This is the equation of simple harmonic motion, and if we put in it the initial conditions $Y(0) = -4.4$ and $\dot Y(0) = 0$ (because $\dot Y = v_2(t) - v_1(t)$ and at $t=0$, $v_1 = v_2$) we can get
$$ Y(t) = x_2(t) - x_1(t) = -4.4\cos\omega t $$
Combining this together with $X(t) = vt$ we can obtain explicit forms for $x_1(t)$ and $x_2(t)$, and also for the velocities.
Now I can draw a parallel to your approach, and show how it is different. What you have done with the conservation of energy amounts to asking, "what are the velocities at the moment when the energy stored in the spring is released, that is, at the moment when the spring is at its normal length?" In the oscillating system, in one cycle there are two such moments - when the spring reaches zero elongation while getting longer (which happens at $t_1 = T/4 = \pi/2\omega$, and while getting shorter (at $t_3 = 3T/4 = 3\pi/2\omega$, after having stretched to a maximum positive elongation at $t_2 = T/2 = \pi/\omega$). The two roots you have found correspond to the velocities of the objects at $t_1$ and $t_3$, respectively: both of these satisfy the exact same conservation law equations. Once the first of these happens, I presume what was intended is that the spring, which is not actually attached, "drops" since nothing is pushing on it from the sides any more, and after that each cart, without any forces acting on it, just keeps going with the velocity it had at that moment. So it really is about which happens first, but your approach doesn't contain any explicit information about time, so it is not immediately obvious which is the root you want.
The conservation laws approach is similar to thinking of the process as a collision. But you have to realize that in a collision, there are two sets of solutions satisfying the equations - the values before the collision, and the values after (in your problem, the initial moment is actually in between those - it is the moment of maximum deformation). If you want to be certain to distinguish them, you could write the momentum conservation law as a pair of equations like this
m_1 v_1 - m_1 v &= -J\\
m_2 v_2 - m_2 v &= J
(or if, in general the initial velocities are different, replace $v$ with $u_1$ and $u_2$ accordingly). The quantity $J$ in these equations is the impulse of the force by which the colliding objects interact (it is just an integral of the force over time), so the equal magnitude and opposite signs of the right-hand-sides are an expression of Newton's third law. If you add them up, you get conservation of momentum. But, if you get two roots, one of them will correspond to positive $J$ (meaning the $m_2$ object was pushed/pulled to the right), while the other to negative $J$ ($m_2$ pushed/pulled to the left). We know which one we need here, since while uncompressing, the spring pushes outward.