Answer
{$-5,-4,1,2$}
Work Step by Step
Plug $f(x)=-16$
Therefore, we have $(x^2+3x-2)^2-10(x^2+3x-2)=-16$
This implies $[(x^2+3x-2)-8][(x^2+3x-2)-2]=0$
Now, $(x^2+3x-2)-8=0$ or, $x^2+3x-10=0$
or, $(x+5)(x-2)=0$ or, $x=${$-5,2$}
and $(x^2+3x-2)-2=0$ or, $x^2+3x-4=0$
or, $(x-1)(x+4)=0$ or, $x=${$-4,1$}
Solution set: $x=${$-5,-4,1,2$}
