How to make 2x^2-4x=-3 a perfect square

The original 2x^2 - 4x = -3 becomes 2[ (x - 1)^2 - 1 = -3.

Step-by-step explanation:

Start with 2x^2-4x=-3.  Rewrite this as  2x^2-4x                   = -3, and then as

2(x^2 - 2x                       ) = -3.  

Next, "complete the square" of x^2 - 2x:  Take half of the coefficient (-2) of x (which comes out to -1).  Square this result (obtaining (-1)^2, or just 1.  Add this 1 to x^2 - 2x, and then subtract the same quantity (1):  x^2 - 2x + 1 - 1.

Now rewrite the perfect square x^2 - 2x + 1 as (x - 1)^2, and then subtract that -1 (above):

(x - 1)^2 - 1.

Substituting this for x^2 - 2x in 2(x^2 - 2x                       ) = -3, we get:

2 [ (x - 1)^2 - 1 ]                  = -3, or

2(x - 1)^2 - 2 = - 3, or

2(x -1)^2 = -1

We weren't asked to solve this equation, but may as well do so:

Divide both sides by 2, obtaining (x - 1)^2 = -1/2

Normally, we'd find the square root of both sides, but here we cannot find the square root of the negative number -1/2, unless complex roots are acceptable.

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