Question 141840
Well, the obvious answer is that 4 is a perfect square, while 2 and 6 are not.  Having said that, the process you showed is not the one that I use.


Here's my process given {{{ax^2+bx+c=0}}}


Move the constant term to the right:
{{{ax^2+bx=-c}}}


Divide both sides by the lead coefficient (if it is already 1, you don't have to do anything)
{{{x^2+(b/a)x=((-c)/a)}}}


Divide the coefficient on the 1st degree term by 2, and then square the result.  Add that result to both sides of the equation.
{{{x^2+(b/a)x+(b^2/4a^2)=(b^2/4a^2)+((-c)/a)}}}


Factor the left now that it is a perfect square.
{{{(x+(b/2a))^2=(b^2/4a^2)+((-c)/a)}}}


Take the square root of both sides.
{{{x+(b/2a)=sqrt((b^2/4a^2)+((-c)/a))}}} or {{{x+(b/2a)=-sqrt((b^2/4a^2)+((-c)/a))}}}


Add {{{-b/2a}}} to both sides.
{{{x=-b/2a+-sqrt((b^2/4a^2)+((-c)/a))}}}


Simplify.
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 


Hope that helps.