#### Answer

$x=\left\{ -4,6 \right\}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given equation, $
x^2-2x-24=0
,$ use first the properties of equality to express the equation in the form $x^2\pm bx=c.$ Once in this form, complete the square by adding $\left( \dfrac{b}{2} \right)^2$ to both sides of the equal sign. Then express the left side as a square of a binomial while simplify the right side. Then take the square root of both sides (Square Root Property) and use the properties of equality to isolate the variable.
$\bf{\text{Solution Details:}}$
Using the properties of equality, in the form $x^2+bx=c,$ the given equation is equivalent to
\begin{array}{l}\require{cancel}
x^2-2x=24
.\end{array}
In the equation above, $b=
-2
.$ The expression $\left( \dfrac{b}{2} \right)^2,$ evaluates to
\begin{array}{l}\require{cancel}
\left( \dfrac{-2}{2} \right)^2
\\\\=
\left( -1 \right)^2
\\\\=
1
.\end{array}
Adding the value of $\left( \dfrac{b}{2} \right)^2,$ to both sides of the equation above results to
\begin{array}{l}\require{cancel}
x^2-2x+1=24+1
\\\\
x^2-2x+1=25
.\end{array}
With the left side now a perfect square trinomial, the equation above is equivalent to
\begin{array}{l}\require{cancel}
(x-1)^2=25
.\end{array}
Taking the square root of both sides (Square Root Property), simplifying the radical and then isolating the variable, the equation above is equivalent to
\begin{array}{l}\require{cancel}
x-1=\pm\sqrt{25}
\\\\
x-1=\pm\sqrt{(5)^2}
\\\\
x-1=\pm5
\\\\
x=1\pm5
.\end{array}
The solutions are
\begin{array}{l}\require{cancel}
x=1-5
\\\\
x=-4
\\\\\text{OR}\\\\
x=1+5
\\\\
x=6
.\end{array}
Hence, $
x=\left\{ -4,6 \right\}
.$