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 #x#s and #y#s 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^2#s 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#
