Answer
The solutions are $x=-3$, $x=-1$, $x=2$ and $x=4$
Work Step by Step
$(x^{2}-x)^{2}-14(x^{2}-x)+24=0$
Let $u$ be equal to $x^{2}-x$
If $u=x^{2}-x$, then $u^{2}=(x^{2}-x)^{2}$
Rewrite the original equation using the new variable $u$:
$u^{2}-14u+24=0$
Solve by factoring:
$(u-12)(u-2)=0$
Set both factors equal to $0$ and solve each individual equation for $u$:
$u-12=0$
$u=12$
$u-2=0$
$u=2$
Substitute $u$ back to $x^{2}-x$ and solve for $x$:
$x^{2}-x=12$
$x^{2}-x-12=0$
$(x-4)(x+3)=0$
$x-4=0$
$x=4$
$x+3=0$
$x=-3$
$x^{2}-x=2$
$x^{2}-x-2=0$
$(x-2)(x+1)=0$
$x-2=0$
$x=2$
$x+1=0$
$x=-1$
The solutions are $x=-3$, $x=-1$, $x=2$ and $x=4$
