As we have to solve #x^2-9x=-12# by completing square method
first have alook at LHS, it appears to be of the form
#a^2-2ab+color(red)(b^2)#, where #a# is #x# tp find #b#,
letus write #-9x# as #2xx x xx (-9/2)# #-# an it is apparent that we can use #-9/2# as #b# and therefore for completing square we must add #color(red)(b^2)# i.e. #(-9/2)^2=81/4#.
Hence we can write #x^2-9x=-12# as
#x^2-9x+color(red)(81/4=-12+color(red)(81/4)#
i.e. #(x-9/2)^2=(-48+81)/4=33/4#, which can be written as
#(x-9/2)^2=(sqrt33/2)^2#
or #(x-9/2)^2-(sqrt33/2)^2=0#
Now, we can write #a^2-b^2# as #(a+b)(a-b)# and our equation becomes
#(x-9/2+sqrt33/2)(x-9/2-sqrt33/2)=0#
i.e. either #x=9/2-sqrt33/2# or #x=9/2+sqrt33/2#
