When completing the square:

5x^2 + 10x - 20
5 ( x^2 + 2x) -20 = 0
5 ( x + 1)^2 -15 = 0

Did I do that right? If so I have another question. When it's in this form, -15 is the y-coordinate of the vertex and 2 the x coordinate of the vertex? If so isnt the x cooridinate of the vertex -2 because you flip the sign?

Hmmmm.

5 ( x^2 + 2x) -20 = 0
add 20 to each side.
5 ( x^2 + 2x)=20

divide each side by 5
x^2+ 2x =4
add one to each side to complete the square

x^2+2x+1=5 factor, and subtract five from each side
(x+1)^1-5=0
I think this is different from yours. Check it.

I don't know how you are getting a vertex. Is there an equation at says y is equal to something? I don't understand your questions.

5 x^2 + 10 x - 20 = y

first divide everything by 5, the coef of x

x^2 + 2 x - 4 = y/5
now add 4 to both sides

x^2 + 2 x = y/5 +4
now take half the coef of x, square it, and add to both sides
(2/2)^2 =1
so
x^2 + 2 x + 1 = y/5 + 5
NOW the left is
(x+1)^2 = y/5 + 1

to the right and left of x = -1 where the left is 0
so the x coordinate of the vertex is x = -1

What is y then?
y/5 + 1 = 0
y = -1(5) = -5
so the vertex is at
(-1,-5)

This is how I did this:

5x^2+10x=20 <--- I left that out, sorry
5x^2+10x-20=0
5(x^2+2x)-20=0
5(x^2+2x-1+1)-20+5
5(x+1)^2-15=0

So I was thinking that (-1,-15) was the vertex because it is in the form y=a(x-h)^2+k

??
Does this make sense?

If you just have an equation in x with no y, you just have two points where the left is equal to the right, no parabola, no vertex.

Yes, you correctly completed the square for the quadratic equation 5x^2 + 10x - 20. However, there appears to be a small mistake in your solution. Let's go through the steps again together:

1. Start with the quadratic equation: 5x^2 + 10x - 20.
2. Factor out the common factor of 5 from the first two terms: 5(x^2 + 2x) - 20 = 0.
3. Complete the square by taking half of the coefficient of the x-term (which is 2) and squaring it. Add that value inside the parentheses preceded by a square (2/2)^2 = 1^2 to maintain the equality.
- Then, rewrite the equation, adding and subtracting the value you just calculated: 5(x^2 + 2x + 1 - 1) - 20 = 0.
- Notice that adding and subtracting 1 inside the parentheses doesn't change the equation since you multiplied it by 5.
4. Simplify the equation within the parentheses: 5(x^2 + 2x + 1 - 1) - 20 = 0.
- The terms inside the parentheses can be rewritten as a perfect square: 5((x + 1)^2 - 1) - 20 = 0.
5. Expand the equation: 5(x + 1)^2 - 5 - 20 = 0.
6. Combine like terms: 5(x + 1)^2 - 25 = 0.
7. Move the constant term to the other side: 5(x + 1)^2 = 25.
8. Divide both sides by 5 to isolate the squared term: (x + 1)^2 = 5.
9. Take the square root of both sides to solve for x + 1: x + 1 = ±√5.
10. Solve for x by subtracting 1 from both sides: x = -1 ± √5.

To answer your question about the vertex coordinates: in the equation 5(x + 1)^2 - 15 = 0, the y-coordinate of the vertex is indeed -15. However, the x-coordinate of the vertex is not 2 or -2. You need to set the equation equal to zero to find the x-coordinate of the vertex. In this case, it is x = -1. The x-coordinate of the vertex is the opposite of the term inside the parentheses. So, in this case, the x-coordinate is -1 because the expression is (x + 1).