Answer:

(x-2.5)^2 + (y+1.5)^2/4 = 1.

Step-by-step explanation:

To transform the equation 4x^2+4y^2-20x+12y-2=0 into standard form, we need to complete the square for both the x and y terms.

First, let's deal with the x terms:

4x^2 - 20x = 4(x^2 - 5x)

To complete the square for the x terms, we need to add and subtract (5/2)^2 = 6.25 inside the parentheses:

4(x^2 - 5x + 6.25 - 6.25) = 4[(x-2.5)^2 - 6.25]

Now, let's deal with the y terms:

4y^2 + 12y = 4(y^2 + 3y)

To complete the square for the y terms, we need to add and subtract (3/2)^2 = 2.25 inside the parentheses:

4(y^2 + 3y + 2.25 - 2.25) = 4[(y+1.5)^2 - 2.25]

Putting it all together, we get:

4[(x-2.5)^2 - 6.25] + 4[(y+1.5)^2 - 2.25] - 2 = 0

Simplifying, we get:

4(x-2.5)^2 + 4(y+1.5)^2 = 16

Dividing both sides by 16, we get:

(x-2.5)^2 + (y+1.5)^2/4 = 1

Therefore, the standard form of the equation is:

(x-2.5)^2 + (y+1.5)^2/4 = 1.