I'll give my shot at explaining this as well (it will help me understand this better also). The following is a long ramble (I tend to do that). If it's something you already understand, then just ignore it, since I don't know how much you know.
As symbolipoint said, when we "complete the square," we are really changing the quadratic equation from some form/representation to an equivalent representation mathematicians call standard form. It is important to realize that the goal is to transform the quadratic equation into standard form, and it doesn't really matter how you do it. "Completing the square" is perhaps the easiest method, but never forget that it is really just a tool to get your equation into standard form.
First of all, why would we want to transform the quadratic equation into standard form? One of the forms of the quadratic equation we are often taught is f(x)=y=ax^2+bx+c, but it is not clear from the equation that the graph should be a parabola. We are told by our teachers that the graph is a parabola, but why?
Take an example. Suppose our equation is y=x^2+4x+1. We graph this equation with several points, and the more points we add, the more we see that, indeed, the graph does look like it is a parabola. But why? We don't see this intuitively. Why does one x^2 plus a 4x plus a 1 give me a parabola? What if I had x^2+4x+2? Would that still be a parabola? What if I had 5x^2+33x+43? Would that still be a parabola?
We try several more sample equations, and we find that all equations in the form y=ax^2+bx+c seem to be parabolas. We have a strong suspicion that they are all parabolas, but we still can't see why from the equation. Let's play around with the equation to see if we can make it look like something we would associate with a parabola.
Take y=x^2+4x+1 again. We can't factor this equation. After staring at it form awhile, however, we may get a bright idea. What if we add something to both sides to make the right hand side something that can be factored? Let's make it easy by just making the right hand side a perfect square. We recall that (x+a)^2 = x^2+2ax+a^2. From our equation, it looks like 2a=4 \implies a=2. Then the constant term a^2=4. Right now, we have a 1, but we want to make it a 4. We can add 3 to both sides, however. Then, we would have
y+3 = x^2 + 4x + 4
You should know that we chose the right hand side to be a perfect square completely arbitrarily. But you'll also see that we were lucky that we made this choice, because it will soon enable us to see why we have a parabola.
This is also why the method is called "completing the square," because we are making a perfect square on the right hand side. Now, you might be thinking, my book would have told me to first subtract 1 on both sides. I would then have
y-1 = x^2+4x \implies y-1+4 = x^2+4x+4,
but can you see that this is really the same thing? We are really adding three to both sides. Textbooks will often give you a routine set of steps. They tell you to subtract the constant on both sides, and then add some constant to both sides, but you can always combine both the subtraction and the addition! If you know that you want a perfect square on the right hand side, then you understand the concept. The objective is not to subtract and add, it is to get a perfect square on the right hand side.
Sometimes, people will also tell you this,
y=x^2+4x+1=x^2+4x+(4-4)+1
That is true, because 4-4=0, but if you understand that we are trying to get a perfect square on both sides, you know that what we really want is
y=(x^2+4x+4) - 4+1=(x^2+4x+4)-3
which you should be able to see is the same thing as what we got before, just with the 3 on the other side.
Now, you will see why it was important to get a perfect square. We now have
y+3 = x^2+4x+4 = (x+2)^2 \implies y=(x+2)^2-3
and this is beautiful because we can see that this is exactly the equation for a parabola! If we substitute, for example, z=x+2, we have y=z^2-3 and we see that this is just a parabola shifted down three units.
So you see that the reason we wanted a perfect square on the RHS was so we could collapse it into a single quantity squared (in this case x+2). Then, it is very clear that the shape of the graph is a parabola.
Well, that's fine, you could say, but what about my situation, when the equation is
y=2x^2-4x+7
We again want to make the RHS into some perfect square. We don't want to worry about the 7 right now, so we can, as in your book, move it over to the left hand side, so we have
y-7=2x^2-4x
It is extremely important that you understand this step is optional! The textbook tells you to do it because it makes looking at the RHS much easier. There isn't a random number there that is distracting you from your objective, which is to make a perfect square.
This is difficult, because we have a 2 in the front. However, we notice that if we divide both sides by 2, then we would have
\frac{y-7}{2}=\frac{y}{2}-\frac{7}{2}=x^2+2x
Now, we know what to do! Give it a try.
You should strive to understand the big concepts. Going through this method of completing the square, what you should learn is why an equation
y=ax^2+bx+c
is a parabola. You should also begin to see that every type of equation in that form has an equivalent standard form that you get after completing the square.
Your teacher, your book, and your class will ask you questions such as, find the vertex, etc. These are important, because they show you applications of the standard form, but you should also understand the connection between the form ax^2+bx+c and the standard form a(x-c)^2+d.
