3x^2+6x-9=0 complete the square

You could simplify this work by factoring "3" out of all four terms, as follows:

3 (x^2 + 2x - 3) = 3 (0) = 0

Hold the 3 for later re-insertion. Focus on "completing the square" of x^2 + 2x - 3.

1. Take the coefficient (2) of x and halve it: 2 divided by 2 is 1

2. Square this result: 1^2 = 1

3. Add this result (1) to x^2 + 2x, holding the "-3" for later:

x^2 + 2x

4 Subtract (1) from x^2 + 2x + 1: x^2 + 2x + 1 - 3 - 1 = 0,

or x^2 + 2x + 1 - 4 = 0

5. Simplify, remembering that x^2 + 2x + 1 is a perfect square:

(x+1) ^2 - 4 = 0

We have "completed the square." We can stop here. or, we could solve for x: one way would be to factor the left side:

[ (x+1) - 2][ (x+1) + 2]=0 The solutions would then be:

x+1-2=0=> x-1=0, or x=1, and

x+1 + 2 = 0 = > x+3=0, or x=-3. (you were not asked to do this).