#### Answer

$x=\left\{ -5,-1 \right\}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
Express the given equation, $
(x+8)(x-2)=-21
,$ in $ax^2+bx+c=0$ form. Then, express the equation in factored form. Next, use the Zero-Factor Property by equating each factor to zero. Finally, solve each equation.
$\bf{\text{Solution Details:}}$
Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
x(x)+x(-2)+8(x)+8(-2)=-21
\\\\
x^2-2x+8x-16=-21
\\\\
x^2+(-2x+8x)+(-16+21)=0
\\\\
x^2+6x+5=0
.\end{array}
To factor the equation above, find two numbers, $m_1$ and $m_2,$ whose product is $c$ and whose sum is $b$ in the quadratic expression $x^2+bx+c.$ Then, express the factored form as $(x+m_1)(x+m_2)=0.$
In the equation above, the value of $c$ is $
5
$ and the value of $b$ is $
6
.$
The possible pairs of integers whose product is $c$ are
\begin{array}{l}\require{cancel}
\{ 1,5 \}
\\
\{ -1,-5 \}
.\end{array}
Among these pairs, the one that gives a sum of $b$ is $\{
1,5
\}.$ Hence, the factored form of the equation above is
\begin{array}{l}\require{cancel}
(x+1)(x+5)=0
.\end{array}
Equating each of the factors in the expression above to zero (Zero-Factor Theorem), then,
\begin{array}{l}\require{cancel}
x+1=0
\text{ OR }
x+5=0
.\end{array}
Solving each of the equations above results to
\begin{array}{l}\require{cancel}
x+1=0
\\\\
x=-1
\\\\\text{ OR }\\\\
x+5=0
\\\\
x=-5
.\end{array}
Hence, the solutions are $
x=\left\{ -5,-1 \right\}
.$