Why does the discriminant in the Quadratic Formula reveal the number of real solutions? Why does the discriminant in the quadratic formula reveal the number of real solutions to a quadratic equation? 

That is, we have one real solution if
$$b^2 -4ac = 0,$$
we have two real solutions if
$$b^2 -4ac &gt; 0,$$
and we have no real solutions if
$$b^2 -4ac &lt; 0.$$
Think about it geometrically $-$ then compute.

Everyone knows $x^2$ describes a parabola with its apex at $(0,0)$. By adding a parameter $\alpha$, we can move the parabola up and down: $x^2+\alpha$ has its apex at $(0,\alpha)$. Looking at the graph as it moves up and down you immediately see how the number of zeros depends on $\alpha$:


for $\alpha&gt;0$ we have no zeros.
for $\alpha=0$ we have a single zero.
for $\alpha&lt;0$ we have two zeros.


Now we can introduct a second paramter $\beta$ to move the parabola left and right: $(x-\beta)^2+\alpha$ has its apex at $(\beta,\alpha)$. 

Note: we used the fact that given a function $f(x)$, the graph of the function $f(x-\beta)$ looks exactly like the one of $f$ but shifted to the right by an amount $\beta$. 

But of course, shifting a function left and right does not alter the amount of zeros. So it still only depends on $\alpha$. We expand the term a bit:

$$(x-\beta)^2+\alpha=x^2-2\beta x+\beta^2+\alpha.$$

Would your quadratic equation be given in this form, you would immediately see the amount of zeros as describes above. Unfortunately it is mostly given as

$$ x^2+\color{red}px+\color{blue}q=0$$

So instead, you have to look at what parts of the $\alpha$-$\beta$-form above corresponds to these new parameters $p$ and $q$:

$$ x^2\color{red}{-2\beta} x+\color{blue}{\beta^2+\alpha} = 0.$$

So we have $p=-2\beta$ and $q=\beta^2+\alpha$. If we only could extract $\alpha$ from these new parameters, we would immediately see the amount of zeros. But wait! We can!

$$\alpha=q-\beta^2=q-\left(\frac p2\right)^2.$$

This is exactly what you know as (the negative of) the discriminant. 



I used the form $x^2+px+q=0$ and you used $ax^2+bx+c=0$. I hope this is not confusing you. Just divide by $a$ (if $a$ is non-zero):

$$x^2+ \color{red}{\frac ba}x+\color{blue}{\frac ca}=0$$

If you set $p=b/a$ and $q=c/a$ and plug this into my discriminant from above you obtain the one you know:

$$\left( \frac {\color{red}p}2 \right )^2-\color{blue}q = \frac{(\color{red}{b/a})^2}4-\color{blue}{\frac ca}=\frac{b^2}{4a^2}-\frac{4ac}{4a^2} = \frac{b^2-4ac}{4a^2}.$$

Because $4a^2$ is always positive it suffices to look at $b^2-4ac$ as you did in your question.
Say your quadratic equation is
$$
ax^2 + bx +c = 0
$$
and suppose that $a&gt;0$ , so that we can imagine $p(x) = ax^2 + bx +c$ to be a parabola which is open at the top. Let's call $m$ the minimum value of $p(x)$ . Graphically it is just the lowest point of the parabola. 

If you now draw a few examples for parabolas, you will notice that, depending on the sign of $m$, the parabola will cut, touch or never touch the horizontal line ($x$-axis). Depending on the situation, you can see how many roots there are:
\begin{align*}
&amp;\text{if} ~~m&gt;0~~ \text{we don't touch the line $\longrightarrow$ no roots} \\
&amp;\text{if} ~~m = 0~~ \text{we touch the line in one point $\longrightarrow$ one root}
\\
&amp;\text{if} ~~m &lt; 0~~ \text{we cross the line $\longrightarrow$ two roots}
\end{align*}
To find the $x$ coordinate of our minimum $m$, we can use the derivative, since it has to be $0$ there.
$$
0 = p'(x) = 2ax + b  ~~~\Longleftrightarrow ~~~ x = -\frac{b}{2a}
$$
so we get
$$
m 
= p \left( -\frac{b}{2a} \right) 
= a \frac{b^2}{4a^2} - b\frac{b}{2a} +c 
= \frac{-b^2 + 4ac}{4a}
$$
Since we are only interested in the sign of $m$, and since $a&gt;0$ , the sign of $m$ will be the same as the sign of $-b^2 +4ac$ . If we combine this with earlier observation, we have
\begin{align*}
&amp;\text{if} ~~~4ac - b^2 &gt;0~~~ \text{we don't touch the line $\longrightarrow$ no roots} \\
&amp;\text{if} ~~~4ac - b^2 = 0~~~ \text{we touch the line in one point $\longrightarrow$ one root}
\\
&amp;\text{if} ~~~4ac - b^2 &lt; 0~~~ \text{we cross the line $\longrightarrow$ two roots}
\end{align*}