This equation can be solved in a number of ways, but I think this way is the simplest and leads to the fewest errors.
#1.#Evaluate the brackets:
#3(x-2)^2-20=55#
In order to evaluate #(x-2)^2#, we just need to remember that any number squared can just be written as being multiplied by itself.
#3(x-2)(x-2)-20=55#
We can evaluate #(x-2)(x-2)# using the FOIL method
#3(x^2-4x+4)-20=55#
#2.#Expand the brackets:
#3(x^2-4x+4)-20=55#
#3x^2-12x+12-20=55#
#3.# Set the equation equal to #0# and collect like terms:
#3x^2-12x+12-20=55#
#3x^2-12x+12-20-55=0#
#3x^2-12x-63=0#
#4.#Solve for the unknown:
#3x^2-12x-63=0#
The easiest way to solve this is to have a coefficient of #1# for #x^2#, as this will make factorising the equation easier:
#x^2-4x-21=0#
At this stage, we have a simplified equation equal to #0#. In order to find the roots of #x#, we need to find two factors which multiply to give #c# (in this case #-21#) and add to give #b# (in this case #-4#):
#x^2-7x+3x-21=0#
Then we factorise each 'pair' of terms in the equation:
#x(x-7)+3(x-7)=0#
Since we are multiplying #(x-7)# by #x# and by #+3#, we can put these two multipliers in one bracket to make solving the equation easier:
#(x-7)(x+3)=0#
#5.#Set each bracket equal to #0# to get values for #x#:
Since we know that if #a*b=0#, then #a# or #b# must be equal to #0#, we can apply that rule here. But since we don't know which factor equals #0#, we set both equal to #0# and solve for both.
#x-7=0#
#x=7#
#x+3=0#
#x=-3#
#x_1 = 7#
#x_2 = -3#
