Answer
$\color{blue}{\left\{4, 7\right\}}$
Work Step by Step
Add $7$ to both sides to obtain:
$t^2-11t+28=0$
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 equation has $b=-11$ and $c=28$.
Note that $28=-7(-4)$ and $-11= -7+(-4)$.
This means that $d=-7$ and $e=-4$.
Thus, the factored form of the trinomial is: $[t+(-t)][t+(-4)]=(t-7)(t-4)$
The equation above can be written as:
$(t-7)(t-4)=0$
Use the Zero-Product Property by equating each factor to zero.
Then, solve each equation to obtain:
\begin{array}{ccc}
&t-7 = 0 &\text{ or } &t-4=0
\\&t=7&\text{ or } &t=4
\end{array}
Checking:
If t=7:
$7^2-11(7)+21=-7
\\49-77+21=-7
\\-28+27=-7
\\-7=-7$
If $t=4$:
$4^2-11(4)+21=-7
\\16-44+21=-7
\\-28+21=-7
\\-7=-7$
Thus, the solution set is $\color{blue}{\left\{4, 7\right\}}$.
