Quick answer:
#2x^2-2x-2=0#
#2x^2-2x = 2#
#x^2 - x = 1#
#x^2 - x + (1/2)^2= (1/2)^2+ 1#
#(x-1/2)^2 = 5/4#
#x-1/2 = +-sqrt5/2#
#x=(1+-sqrt5)/2#
But what did we do and why?
In order to have a habit for how we do this, let's begin by moving the constant to the other side of the equation. We'll add #2# to both sides to get:
#2x^2-2x = 2#
When completed, the square will be on the left, and it will have the form:
#x^2+-2ax+a^2#,
so the next thing to do is get a #1# (which we won't write) in front of the #x^2#. (In fancy terms, we're going to make the coefficient of #x^2# equal to #1#.)
Multiply both sides of the equation by #1/2# (Remember to distribute the multiplication.)
#1/2(2x^2-2x) = 1/2(2)# now simplify:
#x^2 - x = 1#
Now that we have just #x^2#, we can see that the coefficient of #x# is negative, that tells us that the completed square will look like:
#x^2-2ax+a^2# which we will be able to factor: #(x-a)^2#
We have: #x^2 - x = 1#,
Which we can think of as: #x^2 - 1x = 1#,
so we must have:
#2a = 1#.
And that makes #a=1/2#. Square that and add the result to both sides:
#(1/2)^2 = 1^2/2^2 = 1/4#
#x^2 - x +1/4= 1/4+ 1#,
Now factor on the left (we already know how to factor that! See above.) And add on the right.
#(x-1/2)^2 = 1/4+4/4 = 5/4#
#(x-1/2)^2 = 5/4#
Now use the fact that #n^2 = g# if and only of #n = sqrtg or -sqrtg#
(The square of a number equals a given number if and only if the number is either the positive or negative square root of the given.)
#x-1/2 = +-sqrt(5/4) = +-sqrt5/sqrt4 = +-sqrt5/2#
#x-1/2 = +-sqrt5/2# Add #1/2# to both sides:
#x=1/2 +- sqrt5/2# which we often prefer to write as a single fraction:
#x=(1+-sqrt5)/2#
Note: never let yourself think this is some kind pf weird positive and negative number. It is just a convenient way of writing the two solutions:
#x=(1+sqrt5)/2# and #x=(1-sqrt5)/2#.
