Answer
$x=2$ and $x=1$ satisfy the given conditions
Work Step by Step
$f(x)=2x-5$ $,$ $g(x)=x^{2}-3x+8$ $,$ and $(f\circ g)(x)=7$
To find $(f\circ g)(x)$, substitute $x$ by $g(x)$ in $f(x)$ and simplify:
$(f\circ g)(x)=2(x^{2}-3x+8)-5=2x^{2}-6x+16-5=...$
$...=2x^{2}-6x+11$
Set $2x^{2}-6x+11$ equal to $7$:
$2x^{2}-6x+11=7$
Take $7$ to the left side of the equation:
$2x^{2}-6x+11-7=0$
$2x^{2}-6x+4=0$
Take out common factor $2$ from the left side and take it to divide the right side:
$2(x^{2}-3x+2)=0$
$x^{2}-3x+2=\dfrac{0}{2}$
$x^{2}-3x+2=0$
Solve by factoring:
$(x-2)(x-1)=0$
Set both factors equal to $0$ and solve each individual equation for $x$:
$x-2=0$
$x=2$
$x-1=0$
$x=1$
$x=2$ and $x=1$ satisfy the given conditions
