Multiply out the racket giving:#" "6x^2-5x=6#
Subtract 6 from both sides
#" "6x^2-5x-6=0#
Write as#" "6(x^2-5/6x)-6=0#
Take the square outside the bracket and add the correction constant k
#6(x-5/6x)^2-6+k=0#
Remove the #x# from #-5/6x#
#6(x-5/6)^2-6+k=0#
Multiply the #-5/6# by #(1/2)#
#6(x-5/(12))^2-6+k=0#
'~~~~~~~~~~~~~ Comment ~~~~~~~~~~~~~~~~~~~~~~~~~
Consider the part of #(x-5/12)(x-5/12)#
The #(-5/12)^2# is an introduced value that is not in the original equation so we remove it by subtraction. However, this introduced error is in fact #6(-5/12)^2# due to the 6 outside the brackets
#=>k=(-1)xx6xx(-5/12)^2= -1 1/24 =-25/24#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(brown)(6(x-5/(12))^2-6+k=0)" "->" "color(blue)(6(x-5/(12))^2-6-25/24=0#
#=>6(x-5/(12))^2-169/24=0#
#6(x-5/12)^2=169/24#
#(x-5/12)^2=169/144#
Taking the square root of both sides
#x-5/12=+-sqrt(169)/sqrt(144) = +-13/12#
#x=5/12+-13/12#
#x=+18/12" and " -8/12#
#x=+3/2" and " -2/3#
