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#