A method to derive the quadratic formula I wonder if this is a valid method to derive the quadratic formula.
It is not an optimal method at all, but I am intrigued if it is a genuine way or has any gap.
The method starts from the equation whose coefficients are such that it is not necessary to use the quadratic formula to solve it,
$$(a^2 + b^2)x^2+ 2bcx+c^2 =0$$ becouse we can easily solve for x, having at the right side a quadratic form: $$a^2x^2 = -b^2x^2 - 2bcx - c^2 \implies a^2x^2 = -(bx + c)^2 $$
Obtaining the equation $(a^2 + b^2)x^2+ 2bcx+c^2 =0$ is trivial and the coefficient $a$ is squared so that $(a^2 + b^2)$ is only zero when $a$ or $b$ are zero. Of course $a,b,c \in \mathbb{R}$
First we find the roots of our special equation: $a^2x^2 = -(bx + c)^2 $
Square rooting both sides: $ax = \pm i(bx + c) $ where $ i=\sqrt{-1}$
Solving x we obtain: $ x = \dfrac{\pm ic}{a \mp ib}$
Now we are ready to derive the general quadratic formula. Note that the equation $(a^2 + b^2)x^2+ 2bcx+c^2 =0$ is just a quadratic equation $a'x^2 + b'x +c'$ so we can rearrange the coefficients:
$$a'=(a^2 + b^2)$$
$$b'=2bc$$
$$c'= c^2$$
Let's solve for $a$, $b$ and $c$
$$c'= c^2 \implies c = \pm \sqrt{c'} $$
$$b'=2bc \implies b= \frac{b'}{\pm 2\sqrt{c'}}  $$
$$a'=(a^2 + b^2) \implies a = \pm \sqrt{a' - \frac{b'^2}{4c'} }  $$
And now we substitute $a$, $b$ and $c$ in the above result $ x = \dfrac{\pm ic}{a \mp ib}$ and we obtain:
$$ x = \frac{\pm i \sqrt{c'}}{\pm \sqrt{a' - \frac{b'^2}{4c'} - \mp i \frac{b'}{2\sqrt{c'}}  }}$$
Multiplying top and bottom by $\pm \sqrt{c'}$
$$ x = \frac{\pm i c'}{\pm \sqrt{c'} \sqrt{a' - \frac{b'^2}{4c'} - \mp i \frac{b'}{2}  }}$$
Taking $4c'$ out of the square root and simplifying
$$ x = \frac{\pm 2 i c'}{\pm \sqrt{4a'c' - b'^2} \mp i b'}$$
Multiplying up and down by the conjugate of the denominator:
$$ x = \frac{(\pm 2 i c')(\pm \sqrt{4a'c' - b'^2} \mp i b')}{(\pm \sqrt{4a'c' - b'^2} \mp i b')(\pm \sqrt{4a'c' - b'^2} \pm i b')} \implies$$
$$ \implies x = \frac{(\pm 2 i c')(\pm \sqrt{4a'c' - b'^2} \mp i b')}{4a'c'}$$
Note that $ i \sqrt{4a'c' - b'^2} = \sqrt{b'^2 - 4a'c'}$ and $\pm i * \pm ib' = -b'$ So simplifying:
$$ x = \frac{-b'}{2a'} \pm \frac{\pm \sqrt{b'^2 - 4a'c'} }{2a'}$$
And, we've derived the formula.

[Too long for a comment]
There is another way to derive the quadratic formula. But it is so complicated, it is almost barely usable.

Part 1
We begin with: $$x^2-px+q=0.$$ Let's assume this equation is of the form $A(x-\alpha)(x-\beta)$ for some constant $A$ and roots $\alpha$, $\beta$. So $$x^2-px+q=A(x-\alpha)(x-\beta)=Ax^2-A(\alpha+\beta)x+A\alpha\beta$$
Comparing coefficients, it follows that $A=1$, $\alpha+\beta=p$ and $\alpha\beta=q$. Now, with this information, we may analyse  $\alpha^4+\beta^4$. Recall: $$(m+n)^4=\color{red}{m^4+n^4}+2mn(\color{green}{2m^2}+3mn\color{green}{+2n^2})$$
Let $m=\alpha$ and $n=\beta$. We wish to analyse the expression in red, but first we must evaluate the expression in green. Note: $$(\alpha+\beta)^2=\color{green}{\alpha^2+\beta^2}+2\alpha\beta$$ or, since $p=\alpha+\beta$ and $q=\alpha\beta$ we have: $$p^2=\alpha^2+\beta^2+2q$$ so $$\alpha^2+\beta^2=p^2-2q.$$ So now we have that, if: $(\alpha+\beta)^4=\color{red}{\alpha^4+\beta^4}+2\alpha\beta(2\alpha^2+3\alpha\beta+2\beta^2)$ then $$p^4=\alpha^4+\beta^4+2q(2p^2-q)$$ or $$\color{blue}{p^4}-4q\color{blue}{p^2}-\underbrace{(\alpha^4-2q^2+\beta^4)}_{\large (\alpha^2-\beta^2)^2}=0.$$
So now, we have solved a quadratic. Let's replace $\alpha$ with $u$ and $\beta$ with $v$. We have:

Let $x^2-Px-Q=0.$ If there exist $u$ and $v$ such that $$P=4uv\quad ,\quad Q=(u^2-v^2)^2$$ then $(u+v)^2$ is a root.

The quadratic equation can now be found by solving for $u$ and $v$, and then calculating $(u+v)^2$. The algebra becomes incredibly messy here.

Part 2
Expanding $Q$, we have: $$Q=u^4-2(\color{purple}{uv})^2+v^4=u^4-2\bigg(\color{purple}{\frac P4}\bigg)^2+v^4$$ $$\therefore\qquad u^4+v^4=\frac{P^2+8Q}8.$$ Now this is the equation of a super-ellipse, so we can solve for $u$ and $v$ by recalling the familiar trig identity $\cos^2\vartheta+\sin^2\vartheta=1$. Our equation is equivalent to $$\left(\cfrac{u^2}{\sqrt{\frac{P^2+8Q}8}}\right)^2+\left(\cfrac{v^2}{\sqrt{\frac{P^2+8Q}8}}\right)^2=1$$ or, given some simplification, $$u=\sqrt{\frac 12\cos\vartheta\sqrt{\frac{P^2}2+4Q}}\quad ,\quad v=\sqrt{\frac 12\sin\vartheta\sqrt{\frac{P^2}2+4Q}}$$ And now we ought to find $\vartheta$ such that $uv=\dfrac P4$. We have: $$uv=\frac 12\sqrt{\cos\vartheta\sin\vartheta\bigg(\frac{P^2}2+4Q\bigg)}=\frac P4$$ and, since $2\cos\vartheta\sin\vartheta=\sin(2\vartheta)$, then $$\frac P2=\sqrt{\frac 12\sin(2\vartheta)\bigg(\frac{P^2}2+4Q\bigg)}$$ or $$\sin(2\vartheta)=\frac{P^2}{P^2+8Q}=\lambda$$ or $$\vartheta=\frac 12\arcsin\lambda.$$ And now we use the identites, $$\begin{align}\cos\bigg(\frac 12\arcsin\lambda\bigg)&=\frac 12\Big(\sqrt{1+\lambda}+\sqrt{1-\lambda}\Big)\\ \sin\bigg(\frac 12 \arcsin\lambda\bigg)&=\frac 12\Big(\sqrt{1+\lambda}-\sqrt{1-\lambda}\Big)\end{align}$$ to finally derive (partially in terms of $\lambda$): $$u=\frac 12\sqrt{\frac P{\sqrt{2\lambda}}\Big(\sqrt{1+\lambda}+\sqrt{1-\lambda}\Big)}\quad ,\quad v=\frac 12\sqrt{\frac P{\sqrt{2\lambda}}\Big(\sqrt{1+\lambda}-\sqrt{1-\lambda}\Big)}.$$
Finally: $$(u+v)^2=\Bigg(\frac 12\sqrt{\frac P{\sqrt{2\lambda}}\Big(\sqrt{1+\lambda}+\sqrt{1-\lambda}\Big)}+\frac 12\sqrt{\frac P{\sqrt{2\lambda}}\Big(\sqrt{1+\lambda}-\sqrt{1-\lambda}\Big)}\Bigg)^2$$ $$=\frac P{4\sqrt{2\lambda}}\bigg(\sqrt{\sqrt{1+\lambda}+\sqrt{1-\lambda}}+\sqrt{\sqrt{1+\lambda}-\sqrt{1-\lambda}}\bigg)^2=\frac P{2\sqrt{2\lambda}}\Big(\sqrt{1+\lambda}+\sqrt{2\lambda}\Big)$$ $$=\frac P2\Bigg(1+\sqrt{\frac 12\bigg(1+\frac 1{\lambda}\bigg)}\Bigg)=\frac P2\bigg(1+\sqrt{1+\frac{4Q}{P^2}}\bigg)=\frac{P+\sqrt{P^2+4Q}}2$$
So now we can let $\alpha=\dfrac{P+\sqrt{P^2+4Q}}2$ since that is a root, and we know that $\alpha+\beta=P$.$$\therefore \beta=\frac{P-\sqrt{P^2+4Q}}2$$ and we finally have the form: $\dfrac{P\pm\sqrt{P^2+4Q}}2$. Now, substitute $P=-\dfrac BA$ and $Q=-\dfrac CA$.
