What both solutions correspond to is the completion of the orbit the ball is briefly in while it is in free fall, in both directions. Every body in free fall (ignoring air resistance) is for that time in orbit around the center of gravity of the Earth. It's just the the body can not complete that orbit due to the fact it impacts the ground, and in the time-reversed direction, couldn't have come from that orbit initially because of the ground.
It's not mystery getting in the way of the equations, it's the physical ground.
(There are many other deviations from the highly idealized "orbit the Earth as if it was a stationary body in perfect Newtonian physics" but compared to air resistance you will not be able to witness any of those effects with anything you can throw with your arm from the ground.)
The use of a quadratic to solve the throw is a mathematical model. We say that "the value x describing the when the ball lands must satisfy the quadratic equation F(x)=0" but that does NOT imply the opposite, which is "all x that satisfy F(x)=0 describe a valid motion of the ball."
So when we get two answers, e.g. F(-1)=0 and F(15)=0 for the two points when the ball is at ground level, that means only this: if I had thrown the ball from ground level to follow the same curve land in the same place at x=15, then I would have stood 1m further back when I threw it. It does have physical meaning, but there is nothing curious about the physical meaning.
This throw is symmetrical in time though, in the sense that if I throw the ball with the same speed in the opposite direction starting at x=15 then it will land exactly in my hand. (But the equation here is y=F(x) and not parametrized on time).
The equation we use to describe the motion does not contain a term for the ground being there. That assumption exists outside of the model described by the equation, and you use that assumption after the fact to reject a solution that would otherwise be valid, and describe the movement of the ball that's pulled down by gravity.
In this form, it doesn't really describe a proper orbit, just a trajectory of being pulled down by a constant force. I believe this would correspond to an infinitely-ish heavy object located infinitely far below. The proper equation that gives you an orbital curve has the force of gravity proportional to inverse-square distance and point at the center of the body, which is what makes it possible to describe a circular or elliptical motion this way. Parabolic orbits exist too, but they're interpreted as failed orbital capture - "object is moving so fast that it'll curve around and fly away to infinity before turning around and coming back".
And in all cases, the solutions make physical sense (+/- infinity), on the assumption the trajectory doesn't cross the ground, as there's no term for it there :). If you want, you can describe the ground as another equation (or inequality), and solve the resulting system - it'll then be clear what exactly is it that rejects some of the solutions.