Question 930796
Simple Substitution:  Let {{{u=3x^2+2x}}}.  This gives {{{u^2=u}}},
{{{u^2-u=0}}}
{{{u(u-1)=0}}}
{{{u=0}}}  OR  {{{u-1=0}}};



Back-substituting this means {{{highlight_green(3x^2+2x=0)}}}  OR   {{{highlight_green(3x^2+2x-1=0)}}}
and no further need to square both sides.  Each of these can be solved for x.