How can I find the quadratic formula using calculus? Context: A few years ago, when I was in senior year, I participated in the Student for a Day at my local CEGEP (in Quebec, there are just 5 years of middle and high school.A CEGEP is a school you go to to learn pre-university stuff such as calculus and biology. It's the equivalent of 12th grade in the US). Anyways, I attended a calculus lecture (I had no knowledge of calculus at that time) and I remember the prof proved the quadratic formula using calculus. Today, I'm a second-year math major with more calculus knowledge than the average CEGEP student. I tried re-finding the formula using calculus and did not manage to do it. I also checked online and the only thing I found that was not done by completing the square is a Khan Academy video that doesn't work (the thumbnail is there but the page is inexistant). Do any of you know how to solve it?
 A: Consider the differential equation
$$
\frac{dy}{dx}=2ax+b
$$
where $a,b\in \mathbb{R}$ and with initial condition $y(0)=c$. Separating the variables yields
$$
\int dy=\int (2ax+b)dx
$$
Set $u=2ax+b$ in the right integral to get
$$
\int (2ax+b)dx=\int \frac{1}{2a}u\,du= \frac{u^2}{4a}+C
$$
Reverting the substitution, we get $y=\frac{(2ax+b)^2}{4a}+C$. Now we use the initial condition to find $C=c-\frac{b^2}{4a}$. We thus can write 
$$
y=\frac{(2ax+b)^2}{4a}+c-\frac{b^2}{4a}
$$
Now we want to solve $y=0$ so
$$
\begin{align}
0&=\frac{(2ax+b)^2}{4a}+c-\frac{b^2}{4a}\\
\frac{b^2}{4a}-c&=\frac{(2ax+b)^2}{4a}\\
b^2-4ac&=(2ax+b)^2
\end{align}
$$
Solving for the positive and negative root yields the famous result.
Not the most efficient way to get this formula but it works.
A: We can use calculus, at least, to motivate the choices made when completing the square: The axis of symmetry of the graph of a quadratic function $f(x) = ax^2 + bx + c$, $a \neq 0$, passes through its vertex, and in particular the $x$-value, $x_0$, of the line $\{x = x_0\}$ of symmetry is a critical point of the function:
$$f'(x_0) = 2 a x_0 + b = 0.$$
Rearranging gives $$x_0 = -\frac{b}{2 a},$$ so in the coordinates $uy$, where $x = u + \frac{b}{2a}$, the axis of symmetry is the $y$-axis, $\{u = 0\}$. Substituting and simplifying gives that
$$f(x) = a u^2 + \left(c - \frac{b^2}{4a^2}\right) ,$$
which we can solve for $u$ (and for $x$) simply by rearranging, compute the square roots of the expression for $u^2$, and rewriting in terms of $x$.
