To demonstrate, let's try transforming the equation below into standard form.
4x^2 + 9y^2 + 72y - 24x + 144 = 0
The first thing to do is group xs and ys together
(4x^2 - 24x) + (9y^2 + 72y) + 144
Factor out x^2's and y^2's coefficient
4(x^2 - 6x) + 9(y^2 + 8y) + 144 = 0
Before we continue, let's recall what happens when a binomial is squared
(ax + b)^2 = a^2x^2 + 2abx + b^2
For our problem, we want x^2 - 6x and y^2 + 8y to be perfect squares
For x, we know that
a^2 = 1
=> a = 1, a = -1
2ab = -6
2(1)b = -6
2b = -6
b = -3
=> b^2 = 9
Note that substituting a = -1 will result to the same b^2
Hence, to complete the square, we need to add 9
Meanwhile, for y
a^2 = 1
=> a = 1, a = -1
2ab = 8
2(1)b = 8
2b = 8
b = 4
=> b = 16
Now, let's add our b^2s into the equation.
4(x^2 - 6x) + 9(y^2 + 8y) + 144 = 0
4(x^2 - 6x + 9) + 9(y^2 + 8y + 16) + 144 = 0
We added something into the left-hand side, to retain the "equality", we need to add the same value to the right-hand side (or substract the value again from the left-hand side)
4(x^2 - 6x + 9) + 9(y^2 + 8y + 16) + 144 = 0 + 4(9) + 9(16)
4(x^2 - 6x + 9) + 9(y^2 + 8y + 16) + 144 = 180
=> 4(x - 3)^2 + 9(y + 4)^2 + 144 = 180
=> 4(x - 3)^2 + 9(y + 4)^2 = 180 - 144
=> 4(x - 3)^2 + 9(y + 4)^2 = 36
=> (4(x - 3)^2 + 9(y + 4)^2 = 36)/36
=> (x - 3)^2/9 + (y + 4)^2/4 = 1
