I have seen three approaches to “Completing the Square”, as shown below. Each successfully converts a quadratic equation into vertex form.  Which do you prefer, and why?

### First Approach

This approach moves all terms that are not part of a perfect square to the other side of the equation to get them out of the way. Therefore, it is only useful if you are working with an equation, or are able (and willing) to turn an expression into an equation:

$y=2x^2+12x+10$

$y-10=2x^2+12x$

$\frac{1}{2} y-5=x^2+6x$

$\frac{1}{2} y-5+(\frac{6}{2})^2=x^2+6x+(\frac{6}{2})^2$

$\frac{1}{2}y-5+9=x^2+6x+3^2$

$\frac{1}{2}y+4=(x+3)^2$

$\frac{1}{2}y=(x+3)^2-4$

$y=2(x+3)^2-8$

### Second Approach

This approach keeps everything on the same side of the equation, and is therefore suitable for use when working with expressions. It factors the leading coefficient out of all terms before proceeding:

$y=2x^2+12x+10$

$y=2[x^2+6x\;\;\;\;\;\;\;\;\;+5]$

$y=2[x^2+6x+(\frac{6}{2})^2-(\frac{6}{2})^2+5]$

$y=2[x^2+6x+3^2-3^2+5]$

$y=2[(x^2+6x+3^2)-3^2+5]$

$y=2[(x+3)^2-9+5]$

$y=2[(x+3)^2-4]$

$y=2(x+3)^2-8$

### Third Approach

This approach also keeps everything on the same side of the equation, and is therefore suitable for use when working with expressions or equation. It factors the leading coefficient out of only the two “x” terms:

$y=2x^2+12x+10$

$y=2[x^2+6x]+10$

$y=2[x^2+6x+(\frac{6}{2})^2-(\frac{6}{2})^2]+10$

$y=2[x^2+6x+3^2-3^2]+10$

$y=2[x^2+6x+3^2]-2\cdot 3^2+10$

$y=2[(x+3)^2]-18+10$

$y=2(x+3)^2-8$

### Discussion

From my perspective, each of the above approaches has advantages and disadvantages. I find two of the above approaches appealing, and one less so.  What do you think? Are there other other approaches I should include above?