Question 286994
{{{sqrt(x+6)=x}}}
x+6=x^2
x^2-x-6=0
(x-3)(x+2) = 0
so 3 is one of the roots
Is -2 a usable root?
plug it back in the original
sqrt(-2+6)=-2
sqrt(4)=-2

Is that usable?
(-2)^2=4
(+2)^2=4
does +2=-2? no
 does -2=-2? yes