If $a_n=\left[\frac{n^2+8n+10}{n+9}\right]$, find $ \sum_{n=1}^{30} a_n$ I am working on a problem, assuming high school math knowledge.

Let ${a_n}$ be the sequence defined by $$a_n=\left[\frac{n^2+8n+10}{n+9}\right]\,,$$ where $[x]$ denotes the largest integer which does not exceed $x$. Find the value of $ \sum\limits_{n=1}^{30} a_n$.

I honestly do not understand the text in bold. The answer provided is $445$. Could you please explain what I should be looking at here?
 A: Hint
\begin{align*}
\frac{n^2+8n+10}{n+9}&=\frac{n^2+9n-n-9+19}{n+9}\\
&=n-1+\frac{19}{n+9}\\
\therefore \left\lfloor\frac{n^2+8n+10}{n+9}\right\rfloor&=n-1+\left\lfloor \frac{19}{n+9}\right\rfloor
\end{align*}
As $n$ varies from $1$ to $30$, we will have
\begin{align*}
\left\lfloor\frac{19}{\color{red}{1}+9}\right\rfloor&=\lfloor 1.9\rfloor=1\\
\left\lfloor\frac{19}{\color{red}{2}+9}\right\rfloor&=\lfloor 1.7\rfloor=1\\
\vdots & = \vdots\\
\left\lfloor\frac{19}{\color{red}{11}+9}\right\rfloor&=\lfloor 0.95\rfloor=0\\
\vdots & = \vdots\\
\end{align*}
Hopefully you can take it from here.
A: $[x]$ is the largest integer which does not exceed $x$.
For example, $[12.4]=12$, $[10.995]=10$, $[7]=7$, $[-2.3]=-3$.
For this problem $\displaystyle a_n=\left[\frac{(n-1)(n+9)+19}{n+9}\right]=n-1+\left[\frac{19}{n+9}\right]$.
For $n=1,2,\dots, 10$, $a_n=n$.
For $n=11,12,\dots, 30$, $a_n=n-1$.
A: $$\sum\limits_{n=1}^{30} a_n =\sum\limits_{n=1}^{30}(n-1)+ \sum\limits_{n=1}^{30} [\frac{19}{n+9}]=$$
$$ (1+2+3+...+29) + \sum _1^{10} (1) =435+10=445$$
