Answer
$\{-1,3\}$
Work Step by Step
Recall that $|a|=b \implies a=b$ or $a=-b$.
Thus, using the rule above yields:
$x^2-2x=3\quad$ or $\quad x^2-2x=-3$
Solve each equation.
For $x^2-2x=3$:
Subtract $3$ from both sides then siplify:
$\begin{align*}
x^2-2x-3&=3-3\\
x^2-2x-3&=0\end{align*}$
Rewrite $-2x$ as $-3x+1x$.
$x^2-3x+1x-3=0$
Group the terms.
$(x^2-3x)+(1x-3)=0$
Factor out the GCF in each group.
$x(x-3)+1(x-3)=0$
Factor out $(x-3)$.
$(x-3)(x+1)=0$
Use Zero-Product Property, which states that f $ab=0$, then either $a=0$ or $b=0$ or both are $0$ to obtain:
$x-3=0\quad$ or $\quad x+1 =0$
$x=3\quad $ or $\quad x=-1$
For $x^2-2x=-3$:-
Add $3$ to both sides then simplify to obtain:
$
\begin{align*}
x^2-2x+3&=-3+3\\
x^2-2x+3&=0\end{align*}$
The discriminant $\left(b^2-4ac\right)$ of the quadratic trinomial above is $-8$ therefore the equation has no real number solution.
Hence, the solution set of the given equation is $\{-1,3\}$.
