Answer
$\left\{2, 3\right\}$
Work Step by Step
RECALL:
A trinomial of the form $x^2+bx+c$ can be factored if there are integers $d$ and $e$ such that $c=de$ and $b=d+e$.
The trinomial's factored form will be:
$x^2+bx+c=(x+d)(x+e)$
The given trinomial has $b=-5$ and $c=6$.
Note that $6=-2(-3)$ and $-5= -2+(-3)$.
This means that $d=-2$ and $e=-3$
Thus, the factored form of the trinomial is:
$[x+(-2)][x+(-3)] = (x-2)(x-3)$
The given equation maybe written as:
$(x-2)(x-3)=0$
Use the Zero-Factor Property by equating each factor to zero.
Then, solve each equation to obtain:
$\begin{array}{ccc}
&x-2 = 0 &\text{ or } &x-3=0
\\&x=2 &\text{ or } &x=3
\end{array}$
Thus, the solution set is $\left\{2, 3\right\}$.
