Math
posted by Lena .
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 ycoordinate 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)^15=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+10x20=0
5(x^2+2x)20=0
5(x^2+2x1+1)20+5
5(x+1)^215=0
So I was thinking that (1,15) was the vertex because it is in the form y=a(xh)^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.