Question 896466
<pre>
  {{{abs(X)=6-X^2}}}

We square both sides:

  {{{X^2=(6-X^2)^2}}}

  {{{X^2=36-12X^2+X^4}}}
  
  {{{0=X^4-13X^2+36}}}

  {{{0=(X^2-9)(X^2-4)}}}

  {{{0=(X-3)(X+3)(x-2)(X+2)}}}

We get 3, -3, 2, -2

But we must check for extraneous answers:

  {{{abs(X)=6-X^2}}} 
  {{{abs(3)=6-3^2}}}
  {{{3=6-9}}}
  {{{3=-3}}}

3 is not a solution


  {{{abs(X)=6-X^2}}} 
  {{{abs(-3)=6-(-3)^2}}}
  {{{3=6-9}}}
  {{{3=-3}}}

-3 is not a solution


  {{{abs(X)=6-X^2}}} 
  {{{abs(2)=6-2^2}}}
  {{{2=6-4}}}
  {{{2=2}}}

2 is a solution


  {{{abs(X)=6-(-2)^2}}} 
  {{{abs(-2)=6-(-2)^2}}}
  {{{2=6-4}}}
  {{{2=2}}}

-2 is a solution

So there are only two solutions, 2 and -2.

Edwin</pre>