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
$$b^2 -4ac > 0,$$
and we have no real solutions if
$$b^2 -4ac < 0.$$

Because for $a\neq0$ we obtain:
$$ax^2+bx+c=a\left(x^2+\frac{b}{a}x+\frac{c}{a}\right)=a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a^2}\right).$$
Now, we see that if $b^2-4ac<0$ then $\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a^2}>0$,
which says that the equation $ax^2+bx+c=0$ has no solutions.
For $b^2-4ac=0$ we have one root only:
$$x_1=-\frac{b}{2a}$$ and for $\Delta=b^2-4ac>0$ our equation has two distinct roots:
$$x_1=\frac{-b+\sqrt{\Delta}}{2a}$$ and
$$x_2=\frac{-b-\sqrt{\Delta}}{2a}$$ because in this case we obtain:
$$ax^2+bx+c=a\left(x^2+\frac{b}{a}x+\frac{c}{a}\right)=a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a^2}\right)=$$
$$=a\left(x+\frac{b}{2a}-\frac{\sqrt{\Delta}}{2a}\right)\left(x+\frac{b}{2a}+\frac{\sqrt{\Delta}}{2a}\right)=$$
$$=a\left(x-\frac{-b+\sqrt{\Delta}}{2a}\right)\left(x-\frac{-b-\sqrt{\Delta}}{2a}\right)=a(x-x_1)(x-x_2).$$
