#### Answer

$x=\left\{ -4,3 \right\}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
Express the given equation, $
x^2+x-12=0
,$ in factored form. Then, use the Zero-Factor Property by equating each factor to zero. Finally, solve each equation.
$\bf{\text{Solution Details:}}$
The factored form of the equation above is
\begin{array}{l}\require{cancel}
(x-3)(x+4)=0
.\end{array}
Equating each factor to zero (Zero-Factor Property), then
\begin{array}{l}\require{cancel}
x-3=0
\text{ OR }
x+4=0
.\end{array}
Using the properties of equality to solve each of the equation above results to
\begin{array}{l}\require{cancel}
x-3=0
\\\\
x=3
\\\\\text{ OR }\\\\
x+4=0
\\\\
x=-4
.\end{array}
Hence, the solutions are $
x=\left\{ -4,3 \right\}
.$
$\bf{\text{Supplementary Solution:}}$
To factor the expression, $
x^2+x-12
,$ 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 $(x+m_1)(x+m_2).$
In the given expression, the value of $c$ is $
-12
$ and the value of $b$ is $
1
.$
The possible pairs of integers whose product is $c$ are
\begin{array}{l}\require{cancel}
\{ 1,-12 \}, \{ 2,-6 \}, \{ 3,-4 \},
\{ -1,12 \}, \{ -2,6 \}, \{ -3,4 \}
.\end{array}
Among these pairs, the one that gives a sum of $b$ is $\{
-3,4
\}.$ Hence, the factored form of the expression above is
\begin{array}{l}\require{cancel}
(x-3)(x+4)
.\end{array}