how to find upper bound and lower bound of quadratic 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,
$$
1 \leq \frac{f(n)}{20n^2} \leq 1 + O\left(\frac{1}{n}\right).
$$