#### Answer

$\color{blue}{\left\{-3, 7\right\}}$

#### 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 trinomial in the given equation has $b=-4$ and $c=-21$.
Note that $-21=-7(3)$ and $-4= -7+3$.
This means that $d=-7$ and $e=3$
Thus, the factored form of the trinomial is: $[x+(-7)](x+3) = (x-7)(x+3)$
The given equation maybe written as:
$(x-7)(x+3)=0$
Use the Zero-Factor Property by equating each factor to zero.
Then, solve each equation to obtain:
\begin{array}{ccc}
&x-7 = 0 &\text{ or } &x+3=0
\\&x=7 &\text{ or } &x=-3
\end{array}
Thus, the solution set is $\color{blue}{\left\{-3, 7\right\}}$.