Answer
$x=1$ and $x=-\dfrac{4}{3}$ satisfy the given conditions
Work Step by Step
$f(x)=1-2x$ $,$ $g(x)=3x^{2}+x-1$ and $(f\circ g)(x)=-5$
To find $(f\circ g)(x)$, substitute $x$ by $g(x)$ in $f(x)$ and simplify:
$(f\circ g)(x)=1-2(3x^{2}+x-1)=1-6x^{2}-2x+2=...$
$...=-6x^{2}-2x+3$
Set $(f\circ g)(x)$ equal to $-5$:
$-6x^{2}-2x+3=-5$
Take $5$ to the left side:
$-6x^{2}-2x+3+5=0$
$-6x^{2}-2x+8=0$
Take out common factor $-2$ from the left side and take it to divide the right side:
$-2(3x^{2}+x-4)=0$
$3x^{2}+x-4=\dfrac{0}{-2}$
$3x^{2}+x-4=0$
Solve by factoring:
$(x-1)(3x+4)=0$
Set both factors equal to $0$ and solve each individual equation for $x$:
$x-1=0$
$x=1$
$3x+4=0$
$3x=-4$
$x=-\dfrac{4}{3}$
$x=1$ and $x=-\dfrac{4}{3}$ satisfy the given conditions