<s>[INST] On the number of possible solutions for a quadratic equation. Solving a quadratic equation will yield two roots:
$$\frac{-\sqrt{b^2-4 a c}-b} {2 a}$$
and:
$$\frac{\sqrt{b^2-4a c}-b}{2 a}$$
And I've been taught to answer it like:
$$\frac{\pm\sqrt{b^2-4a c}- b}{2 a}$$
Why does it yields only two solutions? Aren't there infinite solutions for that? Is there a proof on the number of possible solutions for a quadratic equation?
 [/INST] A more important question: Why should there be infinitely many solutions?
The roots/solutions of a quadratic equation $ax^2 + bx + c = 0$ are given by looking at when the graphs $y = ax^2 + bx + c$ cuts the $x$-axis. Remember that $a$, $b$ and $c$ are all fixed numbers, e.g. $y = x^2 + 2x + 1.$  Ask yourself this: Why should there be infinity many values of $x$ for which $x^2 + 2x + 1 = 0$? In fact, there is only one single value of $x$ for which $x^2 + 2x + 1 = 0$. This is because $x^2 + 2x + 1 = (x+1)^2$ and so $x = -1$ is the only solution. Find me another value of $x$ for which $x^2 + 2x + 1 = 0$ and I will give you US$1,000,000.
Draw yourself some sketches. Draw the one-parameter family of parabolae given by $y = x^2 - k$ where $k$ is a number we're going to play with. When $k > 0$, the parabola cuts the $x$-axis at two distinct points: $x = \pm \sqrt{k}$. When $k = 0$, the parabola cuts the $x$-axis at one point: $x = 0$. When $k < 0$, the parabola misses the $x$-axis altogether. 
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