The question tells me to use the distance formula to find the equation of a parabola..

I've gotten down the step where I expand..
This is what I have to expand:

(x-2)^2+y^2=(y+7)^2

I'm horrible with expanding, can you help me please?

Bear with me for a minute, as we go back to basics. "Expanding", as you call it, is really just multiplication. Grab a pencil and a piece of paper and multiply two numbers together: 43 x 12. Do it the old fashioned way you learned when you were younger. You should end up with something like this:

\begin{matrix} & & & & 4 & 3\\ & \times & & & 1 & 2\end{matrix}
\overline{\begin{matrix} & & 8 & 6\\ + & 4 & 3 & \end{matrix}}
\overline{\begin{matrix} & & & 5 & 1 & 6\end{matrix}}


Notice that the final step to get the answer is to add a couple of numbers together? There's actually even more addition going on than you usually think about. When we write a 2-digit number like 43, what we're really saying is (40 + 3). The 4 represents the 10's place (hence its real value is 40, not 4), while the 3 represents the 1's place. Ultimately, when we end up adding 86 + 430 to get our final result, we're really saying (80 + 6) + (400 + 30).

So now let's do the same multiplication a little bit differently - the way we'd do it if these were two algebraic binomials that we were "expanding".

43 x 12 = (40 + 3) x (10 + 2)

To "expand" this product, we have to take each number in the first set of parentheses and multiply it by every number in the second set of parentheses, then add them all up. You can think of it like two soccer teams that have to shake hands after a game. Every person on each team has to shake hands with every person on the other team.

For binomials, where there are exactly two things within each set of parentheses, the shortcut most people try to remember is F.O.I.L., which stands for First, Outer, Inner, Last, meaning that you multiply the first number from each set of parentheses together, then you add the product of the outer two numbers (the first number from the first set times the second number of the second set), then the product of the inner numbers, followed by the product of the last numbers from each set.

Let's try it with our example:

(40 + 3) x (10 + 2)
F + O + I + L = (40 x 10) + (40 x 2) + (3 x 10) + (3 x 2) = 400 + 80 + 30 + 6 = 516

Not only did we get the same answer as the "old fashioned" way of multiplying, we even ended up adding the same set of numbers together to get there!

Okay, now that we've reviewed the process for plain ol' numbers, let's do the same thing with real algebra like the problem you've posted.

First, let's expand that first term (x-2)^2.

(x-2)^2=\left (x-2 \right ) \left (x-2 \right )=\left (x + (-2) \right ) \left (x + (-2) \right )

F+O+I+L = \left (x\cdot x \right )+\left (x\cdot (-2) \right )+\left ((-2)\cdot x \right )+\left ((-2)\cdot (-2) \right )=x^2 - 2x - 2x + 4 = x^2 -4x +4

Now can you do the same thing for the right side of the equation, (y+7)^2?

Once you've done that, you just have to solve the equation for y. You should find that the y^2 terms cancel out, and you're just left with something of the form y = ax^2 + bx + c (in other words, a parabola).

Can you tell me the final equation you get?