Question 1001250
{{{ax^2+4a^2x-12a^3=0}}}
{{{a(x^2+4ax-12a^2)=0}}}
If {{{a=0}}} , every real number {{{x}}} is a solution.
If {{{a<>0}}} ,
{{{x^2+4ax-12a^2=0}}}
{{{x^2+4ax=12a^2}}}
{{{x^2+4ax+(2a)^2=12a^2+(2a)^2}}}
{{{(x+2a)^2=12a^2+4a^2}}}
{{{(x+2a)^2=16a^2}}}
So,
either {{{x+2a=4a}}}--->{{{x=4a-2a}}}--->{{{x=2a}}} ,
or {{{x+2a=-4a}}}--->{{{x=-4a-2a}}}--->{{{x=-6a}}} .
You could say {{{highlight(system(x=2a,"or",x=-6a))}}} ,
or {{{highlight(x=-2a +- 4a)}}} .