Question 130831
So far, you are ok.  Add -6 to both sides before you square again.  You will end up with the quartic: {{{x^4-12x^2+4x=0}}}, but there is an x in every term, so one of the roots is 0.  For the other three you will have to solve the cubic:
{{{x^3-12x+4=0}}}.  My advice:  Google 'cubic solver'  The first thing that comes up is a thing to plug in the coefficients and solve it.  It also has a link to a description of how it works -- but unless you plan to stay up all night...