First, we need to write this equation in standard form. To do that, we must expand and then regroup the terms.

\begin{equation*}
\begin{aligned}
(t - 3)^2 \amp= 3(9 - t) \\
t^2-6t+9 \amp= 27-3t \\
t^2-3t-18 \amp= 0
\end{aligned}
\end{equation*}

Next, we factor the left side of the equation.

\begin{equation*}
\begin{aligned}
t^2-3t-18 \amp= 0
(t+3)(t-6) \amp= 0
\end{aligned}
\end{equation*}

We then apply the zero-factor principle by setting each factor equal to zero.

\begin{equation*}
t+3=0 \hspace{.5cm}\text{or}\hspace{.5cm}t-6=0
\end{equation*}

Finally, solving each equation, we find

\begin{equation*}
t=-6 \hspace{.5cm}\text{or}\hspace{.5cm}t=6
\end{equation*}

Therefore the solutions are \(-3\) and \(6\text{.}\)