How do you simplify $\dfrac{6t+9{{t}^{2}}-6}{2+3t}$?

Hint: The given expression $\dfrac{6t+9{{t}^{2}}-6}{2+3t}$ has a quadratic polynomial in the numerator, and a linear polynomial in the denominator. The linear polynomial is in the simplified form. So we need to consider the quadratic polynomial in the numerator, and write it in the standard form as $9{{t}^{2}}+6t-6$. Then, we can take the factor of $3$ common from the numerator to get \[3\left( 3{{t}^{2}}+2t-2 \right)\].

Complete step by step solution:
Let us write the expression given in the question as
$\Rightarrow f\left( t \right)=\dfrac{6t+9{{t}^{2}}-6}{2+3t}$
Writing the quadratic polynomial in the numerator in its standard form, we get
$\Rightarrow f\left( t \right)=\dfrac{9{{t}^{2}}+6t-6}{2+3t}$
Taking $3$ common from the numerator, we get
$\Rightarrow f\left( t \right)=3\left( \dfrac{3{{t}^{2}}+2t-2}{2+3t} \right)$
Now, taking $t$ common from the first two terms we get
$\begin{align}
  & \Rightarrow f\left( t \right)=3\left( \dfrac{t\left( 3t+2 \right)-2}{2+3t} \right) \\
 & \Rightarrow f\left( t \right)=3\left( \dfrac{t\left( 2+3t \right)-2}{2+3t} \right) \\
\end{align}$
Splitting the fraction inside the bracket, we get
\[\begin{align}
  & \Rightarrow f\left( t \right)=3\left( t-\dfrac{2}{2+3t} \right) \\
 & \Rightarrow f\left( t \right)=3t-\dfrac{6}{2+3t} \\
\end{align}\]
But since we know that the denominator of an expression can never be equal to zero, so we have
\[\Rightarrow 2+3t\ne 0\]
Subtracting \[2\] from both the sides, we get
$\begin{align}
  & \Rightarrow 2+3t-2\ne 0-2 \\
 & \Rightarrow 3t\ne -2 \\
\end{align}$
Finally, dividing both sides by $3$ we get
\[\begin{align}
  & \Rightarrow \dfrac{3t}{3}\ne \dfrac{-2}{3} \\
 & \Rightarrow t\ne -\dfrac{2}{3} \\
\end{align}\]

Hence, the given expression is simplified as \[3t-\dfrac{6}{2+3t}\] for $t\ne -\dfrac{2}{3}$.

Note: We can also divide the numerator by the denominator of the fraction in the given expression to get the same simplified expression. Also, do not forget to determine the domain of the expression given in the above question. Although not specified in the question, it is necessary to determine the values of the domain for which the given expression is not defined.