Most people know what the quadratic formula is so I won’t post it here (Besides, I don’t know how to properly post formulas in general).
I was wondering if there is an intuitive explanation as to why the quadratic formula is structured the way it is.
On
The development of the quadratic formula is based on solving the quadratic equation in the form $$ax^2+bx+c=0.$$
By completing the square, we have \begin{align} ax^2+bx+c&=0\\ ax^2+bx&=-c\\ x^2+\frac{b}{a}x&=-\frac{c}{a}\\ x^2+\frac{b}{a}x\color{blue}{+\left(\frac{b}{2a}\right)^2}&=\color{blue}{\left(\frac{b}{2a}\right)^2}-\frac{c}{a}\\ \left(x+\frac{b}{2a}\right)^2&=\frac{b^2-4ac}{4a^2}\\ x+\frac{b}{2a}&=\pm\sqrt{\frac{b^2-4ac}{4a^2}}\\ x&=-\frac{b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}\\ x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \end{align}
On
Let the turning point $T$ of a quadratic $Q$ be $(T_x,T_y)$
It's easily derivable that $$Q(T_x\pm\alpha)=a\alpha^2+T_y$$
(note that $Q$ is symmetrical about $x=T_x$)
It then follows that $$Q(T_x+\alpha)=0\implies \alpha=\pm\sqrt\frac{-T_y}{a}$$
and so: $$Q\left(T_x\pm\sqrt\frac{-T_y}{a}\ \right)=0$$
Which is the quadratic formula, check with $T_x=\frac{-b}{2a}, T_y=c-\frac{b^2}{4a}$
On
The formula must define two roots. As a parabola (quadratic function) is symmetric around a vertical axis, the formula must be of the form
$$p\pm q,$$ where $p$ corresponds to the abscissa of the vertex, which does not depend on $c$. Next, we know that there are cases such that there are no roots, so that the formula must lead to an impossibility.
At this stage, it is natural to choose a form
$$p(a,b)\pm\sqrt{q(a,b,c)}$$ such that $q$ is negative when there are no real roots.
We also notice that we can divide the whole equation by $a$ without changing the roots and we have
$$p\left(\frac ba\right)\pm\sqrt{q\left(\frac ba,\frac ca\right)}.$$
Now we cheat a little and establish the formula for the apex, cancelling the first derivative,
$$2ax+b=0$$
which gives us
$$x_v=-\frac12\frac ba=p\left(\frac ba\right).$$
The ordinate of the vertex is thus
$$y_v=a\left(-\frac b{2a}\right)^2+b\left(-\frac b{2a}\right)+c=c-\frac{b^2}{4a}$$ and its sign decides the number of roots. So we can assume that
$$q\left(\frac ba,\frac ca\right)=\frac ca-\frac14\left(\frac ba\right)^2$$ or some expression that takes the same sign.
On
The equation $$ax^2+bx+c=0$$ can be normalized to a simpler form by using a linear change of variable such as
$$x=pt+q.$$
Plugging in the equation, we get
$$ap^2t^2+(2apq+bp)t+aq^2+bq+c=0.$$
Now (WLOG $a>0$) we are free to set
$$\begin{cases}ap^2=1,\\2apq+bp=0\end{cases}$$
and the equation simplifies to
$$\color{green}{t^2-d=0}$$ for some constant $d$.
Obviously the solutions are
$$t=\pm\sqrt{d}$$ and are real for $d>0$.
Hence the solutions in $x$ are a linear function of $\pm\sqrt d$. Solving for the parameters $p, q, d$, you obtain the classical formulas.
Using the factor theorem, suppose that you have an equation of the form $$x^2 + Bx + C = 0.$$ If it has roots $r$ and $s$, then we have $$x^2+Bx+C = (x-r)(x-s).$$ (This comes because you can write $x^2+Bx+C = (x-r)q(x)+t$, where $t$ is a remainder that must be constant by using Long Division, and then when you plug in $r$ for $x$, you get $0$ on the left and $t$ on the right. So in fact you get $x^2+Bx+C=(x-r)q(x)$. Doing the same with $x-s$ gives you the result).
Multiplying out you get $$x^2 +Bx + C = x^2 - (r+s)x + rs.$$
So we have that $r+s = -B$ and $rs=C$. But what we actually want are $r$ and $s$, even though what we know are $B$ and $C$.
Now notice that $$\begin{align*} (r-s)^2 & = r^2 - 2rs + s^2\\ &= (r^2+2rs+s^2)-4rs\\ &= (r+s)^2 - 4rs\\ &= B^2-4C. \end{align*}$$
So $|r-s| = \sqrt{B^2-4C}$. By exchanging $r$ and $s$ if necessary, we may assume that $r\geq s$, so that $r-s\geq 0$. So we have $r-s = \sqrt{B^2-4C}$.
But that means that $2r = (r+s)+(r-s) = -B + \sqrt{B^2-4C}$, and that $2s = (r+s)-(r-s) = -B-\sqrt{B^2-4C}$. Thus, the two roots are $$ r = \frac{-B+\sqrt{B^2-4C}}{2} \qquad\text{and}\qquad s=\frac{-B-\sqrt{B^2-4C}}{2}.$$
Now, if you have an arbitrary quadratic, $$ax^2 + bx+c = 0,\qquad a\neq 0$$ then dividing through by $a$ you get one of the form $x^2+Bx+C$ with $B=\frac{b}{a}$ and $C=\frac{c}{a}$. Plugging into the formulas, you get $$\begin{align*} r &= \frac{-B+\sqrt{B^2-4C}}{2} \\ &= \frac{ -\frac{b}{a} + \sqrt{\frac{b^2}{a^2} - 4\frac{c}{a}}}{2}\\ &= \frac{-\frac{b}{a} + \sqrt{\frac{b^2-4ac}{a^2}}}{2}\\ &= \frac{-\frac{b}{a}+\frac{1}{|a|}\sqrt{b^2-4ac}}{2}. \end{align*}$$ and similarly $$s = \frac{-\frac{b}{a} - \frac{1}{|a|}\sqrt{b^2-4ac}}{2}.$$ Now, if $a\gt 0$, you can just get rid of the absolute value; and if $a\lt 0$, then you just end up exchanging $r$ and $s$, and so we may just ignore the absolute value as long as we keep both roots. We get: $$\begin{align*} r &= \frac{-\frac{b}{a}+\frac{1}{a}\sqrt{b^2-4ac}}{2} &\quad\text{and}\quad s&= \frac{-\frac{b}{a}-\frac{1}{a}\sqrt{b^2-4ac}}{2}\\ &= \frac{-b+\sqrt{b^2-4ac}}{2a} & &= \frac{-b-\sqrt{b^2-4ac}}{2a}. \end{align*}$$ Combining the two you get that the roots are given by $$\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$
These ideas can be used to obtain the Cardano formulas for solving cubics; they lead to solving a quadratic equation in order to “solve” for the roots. And can also be used to deduce the Ferrari formulas for a quartic, which lead to a cubic equation to solve for the roots. However, if you try to do something similar to solve the quintic, you end up obtaining a degree six equation that must be solved... which eventually leads to the proof that it is impossible to obtain a formula similar these for the general quintic or higher degree equations. The ideas that the roots play symmetric roles and that it may be possible to obtain expressions for them in terms of the coefficients eventually leads you to Galois Theory and the study of the symmetries of the roots.