A quantum particle moving from A to B will take every possible path from A to B at the same time If a quantum particle can take an unlimited number of paths to get from point A to point B wouldn't a quantum particle never get from point A to point B? 
A quantum particle takes every path at the same time to get from A to B? How is that even possible? Can anyone really explain what is going on? And maybe shed some light on the math?
 A: A quantum object is not something that can "take a path". It only has a definite position if its position is being measured.
The reason it is said that the quantum object "takes every possible path at once" is that the probability for finding a particle that was at position $q_i$ at time $t_i$ at position $q_f$ at time $t_f$ is given by the Feynman path integral
$$ \langle q_f,t_f \vert q_i,t_i\rangle = \langle q_f \vert \mathrm{e}^{\mathrm{i}H(t_i -  t_f)}\vert q_i \rangle = \int_{q(t_i) = q_i}^{q(t_f) = q_f} \mathrm{e}^{\mathrm{i}S[q]}\mathcal{D}q$$
where $S[q] = \int_{t_i}^{t_f} \left(\dot{q}(t)^2 - V(q(t))\right)\mathrm{d}t$ is the action evaluated on the continuous path $q(t)$ and the integral is over all continuous paths with startpoint $q_i$ and endpoint $q_f$, sometimes also called the conditional Wiener space and $\mathcal{D}q$ together with the canonical kinetic term in $S[q]$ is the (conditional) Wiener measure.
Since the integral is over all paths from $q_i$ to $q_f$ and every $\mathrm{e}^{\mathrm{i}S[q]}$ for every path contributes to the value of the integral (that is, we must not neglect any such possible path1), it is said that the Feynman path integral represents the quantum object "talking all possible paths" from $q_i$ to $q_f$. 
The actual quantum time evolution by $\mathrm{e}^{\mathrm{i}H(t_i - t_f)}$ does not look like this, it just traces out one path in the quantum space of states which has nothing to do with the classical paths.

1As GlenTheUdderboat points out in a comment, this is not strictly true. Just as a usual one-dimensional integral can neglect individual discrete points, here, too, we may neglect a zero measure set. Nevertheless, the path integral is defined as the integral over all (continuous) paths, and there is no distinguished zero measure set of paths which we might want to take out.
A: Let me try a low-tech explanation, expanding on @JonCuster comment.
Think about waves. If you drop a pebble in water, a wave spreads out in a circle.
If there is a short wall in the water, the wave will curve around the wall and go the other way.
If you simultaneously drop two pebbles a small distance apart, each starts its own wave in a circle.
These waves interfere, so in some places they reinforce, and in other places they cancel, but you could say in all places both waves are there, and they are just combining.
Suppose you drop 100 pebbles in a line 1 inch apart.
They all start their own waves, but what do you see?
Out to the sides, the waves pretty much always cancel, and perpendicular to the line of pebbles the waves reinforce along a line that moves away.
At intermediate angles, they partially reinforce.
If you drop 200 pebbles 1/2 inch apart, you'll see that the effect is stronger - the waves at intermediate angles are weaker.
Light is the same way, when you consider it as a wave.
When you shine a laser, you can think of it as a very large number of pebbles all in a line being dropped together, so the wave is very coherent in the forward direction.
It still spreads out a bit at the edges.
Well, the power in that wave at a given place is nothing more the probability of seeing a number of photons in that place, over a given time.
So to try to answer your question, waves take all possible paths, but waves interfere so you only see photons in places where the waves interfere constructively.
You might be tempted to say the photon took all possible paths, but that's not right.
The waves take all possible paths, just as they do with pebbles in water.
A: Let's take a simple thought experiment which involves no quantum effects and is quite easy to visualise.
Consider a table on which I mark two points (on its surface); and I ask you to draw the straight line between these two points.
This is easy to do, you merely take a piece of chalk and by eye draw the straight line between them; a moments work.
But there is an alternative method: you draw every possible line connecting the two points, measure it's length, and then chose the shortest one. This is a lot more work, but is exactly equivalent.
This is a cautionary tale in that one must be judicious in interpreting mathematics in a physical situation.
For example, in your situation there is an alternative interpretation: the de Broglie-Bohmian perspective where a pilot wave is posited, and it is it's motion that 'goes everywhere', so to speak, that is described by the description you've given and it acts causally on the particle to guide it - which then follows a classical path.
