How to show that every quadratic, asymptotically nonnegative function $\in \Theta(n^2)$
So I actually found the answers I was looking for. The case $c \ge 0$ is already handled in the question above. For the case $c &lt; 0$ we have:

\begin{align}
n_{1/2} 
&amp;=   \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\
&amp;\le \frac{|b| + \sqrt{b^2 - 4ac}}{a}   \\
&amp;\le \frac{|b| + \sqrt{b^2} + \sqrt{-4ac}}{a} \\
&amp;=   \frac{2|b| + \sqrt{4a|c|}}{a}  \\
&amp;=   \frac{2|b| + 2\sqrt{\frac{a^2|c|}{a}}}{a} \\
&amp;=   2\frac{|b|}{a} +  2\sqrt{\frac{|c|}{a}} \\
&amp;\le 2\max(\{|b|/a, \sqrt{|c|/a}\}) \\
&amp;= n_0
\end{align}

This value for $n_0$ includes the case $c \ge 0$. If $f$ doesn't have any roots, we can instead choose $n_0 = 1$.

For the constants $c_1, c_2$ the authors gave us the values $c_1 = a/4$ and $c_2 = 7a/4$. To check that these are correct we do the following:

Since $n \ge 2|b|/a$ and $n \ge 2\sqrt{|c|/a}$, we know that:
\begin{alignat}{3}
&amp;&amp; \frac{1}{2} &amp;\ge  &amp;\frac{|b|}{an} \quad&amp;\text{and}\quad &amp;\frac{1}{4} \ge  &amp;\frac{|c|}{an^2} \\
&amp;&amp;-\frac{1}{2} &amp;\le -&amp;\frac{|b|}{an} \quad&amp;\text{and}\quad -&amp;\frac{1}{4} \le -&amp;\frac{|c|}{an^2}
\end{alignat}

This gives us

\begin{alignat}{2}
&amp;\frac{1}{4} =   1 - \frac{1}{2} - \frac{1}{4} \le 1 - \frac{|b|}{an} - \frac{|c|}{an^2} \le 1 + \frac{b}{an} + \frac{c}{an^2} \\
\text{and therefore}\quad
&amp;\frac{a}{4}n^2 \le an^2 + bn + c
\end{alignat}

and 

\begin{alignat}{2}
&amp;1 + \frac{b}{an} + \frac{c}{an^2} \le 
 1 + \frac{|b|}{an} + \frac{|c|}{an^2} \le
 1 + \frac{1}{2} + \frac{1}{4} =
 \frac{7}{4} \\ 
\text{and therefore}\quad
&amp;an^2 + bn + c \le \frac{7a}{4}n^2
\end{alignat}

which shows that the values $c_1 = a/4$ and $c_2 = 7a/4$ are indeed correct.