how to find upper bound and lower bound of quadratic equation I am relatively new to algorithms, I wrote one pattern matching algorithm and its running time is $O(n^2)$, I tried it by step count method, direct method and also the constant method which all yields same results. Anyhow now I am writing the upper and lower bound for that, the equation which I get is

$$20n^2+11n+1$$
I need to find the 

$$c_1n^2 \leq 20n^2+11n+1 \leq c_2n^2$$

Here I am stuck and needs some assistance, I have gone through a lot of examples, books, and youtube videos, but no one has any method to solve it instead of assumption and different ways. Any help is much appreciated.

EDIT
Graph for the equation
Since $n \geq 1$ (since $n$ is the input length), we have
$$
20n^2 \leq 20n^2 + 11n + 1 \leq (20+11+1)n^2 = 32n^2.
$$
Put differently, for all $n \geq 1$, the function $f(n) = 20n^2 + 11n + 1$ is bounded from below by $20n^2$ and from above by $32n^2$. This shows that $f(n) = \Theta(n^2)$.

Note also that when $n$ is large, $f(n)$ gets very close to $20n^2$. Indeed,
$$
1 \leq \frac{f(n)}{20n^2} \leq 1 + O\left(\frac{1}{n}\right).
$$