Answer
$\left\{-3, 2\right\}$
Work Step by Step
RECALL:
A quadratic trinomial $x^2+bx+c$ may be factored if there are integers $d$ and $e$ such that $de = c$ and $d+e =b$
If such integers exist, then
$x^2+bx + c = (x+d)(x+e)$
The trinomial in the given equation has $c=-6$ and $b= 1$.
Note that $-6=3(-2)$ and $3+(-2) = 1$
This means that $d=3$ and $e = -2$ and the given trinomial may be factored as:
$x^2+x-6
\\=(x+d)(x+e)
\\=(x+3)[x+(-2)]
\\=(x+3)(x-2)$
Thus, the given equation is equivalent to:
$(x+3)(x-2) =0$
Use the Zero-Product Property by equating each factor to zero then solving each equation to obtain:
$\begin{array}{ccc}
&x+3=0 &\text{ or } &x-2=0
\\&x=-3 &\text{ or } &x=2
\end{array}$
Therefore the solution set is $\left\{-3, 2\right\}$.
