TITLE: How do you derive the quadratic formula using calculus?
QUESTION [11 upvotes]: The quadratic formula: $$f(x)=ax^2+bx+c=0$$
$$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$
I remember a tutor once showing me a method for deriving the quadratic formula using calculus somehow. This was around 20 years ago and I can't even remember the tutor's name. I'd really like to learn this method. Just to clarify, I do know how to derive it using the "Completing the square" method.
I was linked to the solution here: https://www.google.com/amp/s/threesixty360.wordpress.com/2008/10/19/using-calculus-to-generate-the-quadratic-formula/amp/
But I am stuck at one step.
Start with: $$f(x)=ax^2+bx+c$$
We want: $$f(x)=0$$
The first derivative gives: $$f'(x)=2ax+b$$
Which leads to this: $$f(x)=c+\int_0^x (2at+b)dt$$
I can't see why the $t's$ were introduced here.
If anyone has any other methods I'd really like to see them also.

REPLY [3 votes]: It seems what's confused you is the fact that, whereas an indefinite integral uses the same variable on both sides of an equation such as $\int x^2dx=\frac13x^3+C$, you can't use $x$ as both a definite integral's limit and its integration variable, i.e. you have to write $f(x)=f(0)+\int_0^xf^\prime(t)dt$ instead of $f(x)=f(0)+\int_0^xf^\prime(x)dx$. I'm sure you can follow the rest of the proof; it's just an overly complicated way of deducing $f=0$ is equivalent to a condition of the form $w^2=k$. You may want to try substituting $v=t+b/2a$ instead to check you understand the technique.