Question 909908
removing the absolute vale leads to

case 1:
{{{x^2+3x=x^2-2x}}}  or 

case 2:
{{{x^2+3x=-(x^2-2x)}}}  


case 1:


{{{3x=-2x}}}

{{{5x=0}}}

{{{x=0}}}


case 2:

{{{x^2+3x=-x^2+2x}}}

{{{2x^2+x=0}}}

{{{x=0}}}  or  {{{x=-1/2}}}


solutions:  {0, -1/2}