Question

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

Answer 1

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).