I understand how the midpoint of a parabola is found when one of the roots is $0$ ( it is just $-b/2a$ )
but I can't understand why $\sqrt{\Delta} $ in the quadratic formula $=$ the distance between any of our real roots and our midpoint.
I want to know the proof behind this
On
If the roots are $r_1,r_2$, then the sum of the roots $r_1+r_2=\frac{-b}{a}$ and the product of the roots is $r_1r_2=\frac{c}{a}$.
The distance between the two (real) roots is given by $$|r_1-r_2|=\sqrt{(r_1+r_2)^2-4r_1r_2}.$$
The distance between the mid-point and either of the roots is $\frac{|r_1-r_2|}{2}$. Now you can fill in the details.
On
We know that a parabola is of the form $f(x) =ax^2+bx+c=0$. We can find the roots of the equation by using the quadratic formula: $$ r_{1,2} = \frac{-b \pm \sqrt{b^2-4ac}}{2a} = \frac{-b \pm \sqrt{\Delta}}{2a} $$ Suppose that a parabola $\mathscr{P}$ has two roots $r_1, r_2$ then the midpoint of the parabola is $x_1 = -\frac{b}{2a}$, which is also the axis of symmetry of the parabola. To see this, we rewrite the parabolaformula as: $$ f(x) = a(x-e)^2 +k $$ where $$ e = -\frac{b}{2c}, \:\:k= \frac{4ac-b^2}{4a} $$ (you can check for yourself that these are equivalent). Now it is clear, from the square, that $f$ is symmetric around the point $e = -\frac{b}{2a}$.
Because of the symmetry, the distance from the roots to $x_{\text{1}}$ must be constant.
We are considering a parabola with two roots, thus $\Delta>0$. The roots are then: $$ r_1 = \frac{-b + \sqrt{\Delta}}{2a}, \:\: r_2 = \frac{-b - \sqrt{\Delta}}{2a} $$ so the distance from the midpoint $x_1$ to the roots is: \begin{align*} d_1 &= \left|x_1 - r_1 \right| = \left| - \frac{b}{2a} - \frac{-b + \sqrt{\Delta}}{2a}\right| = \frac{\sqrt{\Delta}}{2|a|} \\ d_2 &= \left| x_1 - r_2 \right| = \left| - \frac{b}{2a} - \frac{-b - \sqrt{\Delta}}{2a} \right| = \frac{\sqrt{\Delta}}{2|a|} \end{align*} so the distance between the roots and the midpoint is, in both cases, equal to $\frac{\sqrt{\Delta}}{2a}$
Assuming what you are interested in is the derivation of $\Delta$ and a justification for its role as the distance between the solutions I think this might be helpful.
Consider the quadratic equation $ax^2 + bx + c = 0$ where $a,b,c \in \mathbb{R}$ and $a \neq 0$. Then, one can factor the $a$ out to obtain $a(x^2 + \frac{b}{a}x) + c = 0$. Since $a \neq 0$ one can also divide both sides of the equation by $a$ to obtain $(x^2 + \frac{b}{a}x) + \frac{c}{a} = 0$. The first non trivial manipulation is the recognition that $(x^2 + \frac{b}{a}x)$ is fairly close to $(x + \frac{b}{2a})^2$. Indeed, there is only a term of $\frac{b^2}{4a^2}$ of difference. Therefore, one can write
\begin{align} (x^2+\frac{b}{a}x) + \frac{c}{a} &= (x + \frac{b}{2a})^2 -\frac{b^2}{4a^2} + \frac{c}{a}. \end{align} Rearranging, one obtains \begin{equation} (x + \frac{b}{2a})^2 = \frac{b^2 - 4ac}{4a^2}. \end{equation} The numerator of the term on the right can be recognised as $\Delta$. Thus, \begin{equation} (x + \frac{b}{2a})^2 = \frac{\Delta}{4a^2}. \end{equation} The last step is to take square roots of both sides to find $x$ alone. However, we know that there are two real numbers such that their square equals $\Delta$ and that both these are solutoins. Thus, we find the usual quadratic formula \begin{equation} x = \frac{-b \pm \sqrt{\Delta}}{2a}. \end{equation} or more suggestively \begin{equation} x = \frac{-b}{2a} \pm \frac{\sqrt{\Delta}}{2a}. \end{equation} From this expression it is clear that the roots of the equations are going to be on either side of $\frac{-b}{2a}$ one being $\frac{\sqrt{\Delta}}{2a}$ towards one side and the other being $\frac{\sqrt{\Delta}}{2a}$ towards the other side of the vertex.