We have the following quadratic equation in our problem:
#color(red)(2x^2 - 56x + 6 = 0)# #color(blue)(...Equation.1)#
Note that the coefficient of #x^2 term# is greater than 1
#color(green)(Step.1)#
In our #color(blue)(...Equation.1)#, we are going move the constant term from the left hand side(LHS) to the right side(RHS).
The constant term is #+6# in our #color(blue)(...Equation.1)#
Add #color(blue){(-6)}# to both sides of our equation.
#color(red)(2x^2 - 56x + 6 - 6 = 0-6)#
#color(red)(2x^2 - 56x = -6)# #color(blue)(...Equation.2)#
#color(green)(Step.2)#
Since the coefficient of #(x^2 term)# is greater than 1, divide out every term of our equation by #2#. This process will make it easier to complete the square
So, our #color(blue)(...Equation.2)# will now become:
#color(red)((2x^2)/2 - (56x)/2 = -6/2)#
Simplifying we get,
#color(red)(x^2 - 28x = -3)# #color(blue)(...Equation.3)#
#color(green)(Step.3)#
We are going to add a term to both sides of equation as follows:
#color(red)(x^2 - 28x + square = -3 + square)#
What are we going to write in the box?
Divide the coefficient of #(-28x)# by #2# and square it.
#((-28)/2)^2#
We get, #14^2 = 196#
This value of #196# goes into the box.
Hence, we get
#color(red)(x^2 - 28x + 196 = -3 + 196)#
#rArr color(red)(x^2 - 28x + 196 = 193)# #color(blue)(...Equation.4)#
#color(green)(Step.4)#
We can now write the LHS in #color(blue)(...Equation.4)# as a Perfect Square
Divide the coefficient of the term #-28x# by 2 and use it to write LHS as a perfect square as shown below:
#rArr color(red)((x - 14)^2 = 193)# #color(blue)(...Equation.5)#
#color(green)(Step.5)#
Take the square root on both sides to simplify.
#color(blue)(...Equation.5)# will now become
#color(red)(sqrt((x-14)^2) = +-sqrt(193))#
We notice that on the LHS both the square root and the square will cancel out to yield
#color(red)((x-14) = +-sqrt(193))# #color(blue)(...Equation.6)#
#color(green)(Step.6)#
Using our #color(blue)(Equation.6)# we get two solutions
#color(red)((x-14) = +sqrt(193) and (x-14) = -sqrt(193)#
Hence, rearranging the terms, our final solutions are
#color(blue)(x = 14 + sqrt(193), x = 14 - sqrt(193)))#
I hope this helps.
