Answer
{$-4,-3,1,2$}
Work Step by Step
Plug $f(x)=-6$
Therefore, we have $(x^2+2x-2)^2-7(x^2+2x-2)=-6$
This implies $[(x^2+2x-2)-6][(x^2+2x-2)-1]=0$
Now, $(x^2+2x-2)-6=0$
or, $x^2+2x-8=0$
or, $(x+4)(x-2)=0$
or, $x=${$-4,2$}
and $(x^2+2x-2)-1=0$
or, $x^2+2x-3=0$
or, $(x+3)(x-1)=0$
or, $x=${$-3,1$}
Solution set: $x=${$-4,-3,1,2$}
