Let's start with the basics. Force, measured in Newtons, can be broken down further into units of mass * acceleration. So...

As far as physics goes, this is pretty simple so far. So, what happens if you apply a force for a period of time, say, one newton for one second?

We end up with units of mass and velocity, in kg and m/s. This means that a force multiplied by a time is identical to a mass multiplied by a velocity. If you know any of the three, you can use algebra to solve for the fourth. In other words,

Instead, though, we'll call the velocity "Δv," with the Greek character delta representing change. This is because a force doesn't set the velocity, it changes the velocity based on what the initial velocity was. So,

Next, we'll move the mass over to the left side, so that we have isolated Δv. The Δv is what we're looking for, because that's how fast a rocket can go. Generally speaking, faster is better... yes? Right! Of course.

But, with rockets, the mass can change by a significant amount. Most of a rocket's mass comes from its fuel supply, so it isn't constant. It changes. Considering, we're going to need to edit this equation a little to make it more useful.

We can say that's the instantaneous delta v, but what we're really looking for is the total delta v. So, let's break this up into small pieces of time, and sum that up over a variable range of m.

Now we just need an equation for m. Let's say that the starting mass is m₀, and the mass decreases at a rate of ṁ. So, this leads to:

Note that we could also consider ṁ the rate of fuel consumption by mass. Substituting this back into the last equation, we get:

The evaluation of this integral from t = 0 to t = t results with:

Since ṁ is the rate of fuel consumption and t is the time of the burn, ṁ * t is the total amount of fuel burned. Since mass can never be negative, we can also eliminate the absolute value signs. Considering, we can rewrite this as:

Where m_f is the final mass, dry mass, or the mass of the rocket after the burn. This can be further simplified using properties of logarithms.

And this is identical to the "delta v" equation on the KSP wiki. Hooray! I did something today!