How do you derive the quadratic formula using calculus? 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.
 A: If f is a polynomial of degree $n$ then for all $x, y$ we have $$f(x)=\sum_{j=0}^n(x-y)^jf^{(j)}(y)/j!$$ where $f^{(0)}=f$ and $f^{(j)}$ is the $j$th derivative of $f$ when $j>0.$... And with the usual convention that $0^0=1 $ (i.e. the term $(x-y)^j$ for $j=0$, when $x=y$).
When $f(x)=Ax^2 +Bx+C$ with $A \ne 0,$ then $f'(x)=2Ax+B$ is equal to $0$ when $x=x_0=-B/2A.$  For all $x$ we have $$f(x)=f(x_0)+(x-x_0)f'(x_0)+(x-x_0)^2f''(x_0)/2!.$$ But $f'(x_0)=0$ and $f''(x_0)=2A,$ so for all $x$ we have $$f(x)=f(x_0)+(x-x_0)^2\cdot A.$$  This  "completes the square" for us.
A: 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.
A: Let 
$$
y=ax^2+bx+c \tag{1}
$$ 
and $\tfrac{dy}{dx}=2ax+b$. Now let $g=\tfrac{dy}{dx}$ such that
$$
\begin{align*}
g &=2ax+b \tag{2} \\
x &= \dfrac{g-b}{2a} \tag{3}.
\end{align*}
$$
Substitute $(3)$ into $(1)$ and get
$$
\begin{align*}
y &=a {\left( \dfrac{g-b}{2a} \right)}^2+b {\left( \dfrac{g-b}{2a} \right)}+c \\
&= a \dfrac{g^2-gb+b^2}{4a^2} + \dfrac{bg-b^2}{2a}+c \\
&= \dfrac{g^2-gb+b^2}{4a} + \dfrac{2bg-2b^2}{4a}+ \frac{4ac}{4a} \\
y &= \dfrac{g^2-b^2 +4ac}{4a} \\
\end{align*}
$$
Set $y=0$ and solve for $g$.
$$
\begin{align*}
y &= \dfrac{g^2-b^2 +4ac}{4a} \\
0 &= \dfrac{g^2-b^2 +4ac}{4a} \\
0 &= g^2-b^2 +4ac \\
g &= \pm \sqrt{b^2-4ac} \tag{4} \\
\end{align*}
$$
Now substitute $(2)$ in for $g$ and solve fo $x$.
$$
\begin{align*}
g &= \pm \sqrt{b^2-4ac} \tag{4} \\
2ax+b &= \pm \sqrt{b^2-4ac}  \\
x &= \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}.
\end{align*}
$$
Another way is: let $f(x)=ax^2+bx+c$ such that
$$
f(x)=\int_{0}^{x} \left( 2at+b \right) dt + c
$$
The $t$'s are introduced here because we cannot have $x$ as the variable of integration and also as a bound. Think of them as dummy variables; all the $t$'s will become $x$'s when you evaluate the integral.
Introduce the "u" substitution where $u=2at+b \implies dt= \tfrac{1}{2a}du$. 
$$
f(x)=\int_{t=0}^{t=x} \dfrac{u}{2a} du + c
$$
Since we are now integrating with respect to $u$, so your new lower and upper bounds respectively become $u=b$ and $u=2ax+b$. Thus, you get
$$
\begin{align*}
f(x) &=\int_{b}^{2ax+b} \dfrac{u}{2a} du + c \\
& = \dfrac{1}{2a} \left( {\dfrac{ {\left( {2ax+b} \right)}^2}{2} - \dfrac{b^2}{2}} \right) + c\\
& = \dfrac{{\left( 2ax+b \right)}^2 }{4a} - \dfrac{b^2}{4a} + \dfrac{4ac}{4a}. \\
\end{align*}
$$
Set $f(x)=0$ and solve for $x$.
$$
\begin{align*}
0 & = \dfrac{{\left( 2ax+b \right)}^2 }{4a} - \dfrac{b^2}{4a} + \dfrac{4ac}{4a}. \\
0 & = \left( 2ax+b \right)^2 - b^2 + 4ac \\
\left( 2ax+b \right)^2 & = b^2 - 4ac \\
x &= \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\
\end{align*}
$$
A: Let $\alpha,\beta$ be the two roots. Then we have by comparing coefficients$$\alpha+\beta=\frac{-b}{a}$$ and also you get $$\alpha.\beta=\frac{c}{a}$$. From here you immediately get $(\alpha-\beta)^2=(\frac{b}{a})^2-4.\frac{c}{a}$ and then you solve for $\alpha, \beta$ 
Note: $(\alpha+\beta)^2=(\alpha-\beta)^2+4.\alpha.\beta$
