Where does the plus-minus come from in the quadratic formula? In the formula
$$x_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
for solving quadratic equations, where does the $\pm$ come from?
The square root already results in both a positive and negative term, is the $\pm$ not therefore extraneous?
 A: What about completing squares?
\begin{align*}
ax^{2} + bx + c & = a\left(x^{2} + \frac{bx}{a}\right) + c\\\\
& = a\left(x^{2} + \frac{bx}{a} + \frac{b^{2}}{4a^{2}}\right) + c - \frac{b^{2}}{4a}\\\\
& = a\left(x + \frac{b}{2a}\right)^{2} + \frac{4ac - b^{2}}{4a}
\end{align*}
Hence we deduce that
\begin{align*}
ax^{2} + bx + c = 0 & \Longleftrightarrow a\left(x + \frac{b}{2a}\right)^{2} + \frac{4ac - b^{2}}{4a} = 0\\\\
& \Longleftrightarrow \left(x + \frac{b}{2a}\right)^{2} = \frac{b^{2} - 4ac}{4a^{2}}\\\\\
& \Longleftrightarrow x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a} 
\end{align*}
and we are done.
Hopefully this helps!
A: Given a general quadratic
\begin{eqnarray*}
ax^2+bx+c=0.
\end{eqnarray*}
Multiply through by $4a$, add $b^2$ & this will allow us to "complete the square"
\begin{eqnarray*}
4a^2x^2+4abx+b^2 &=& b^2-4ac\\
(2ax+b)^2 &=& b^2-4ac.\\
\end{eqnarray*}
Now take the square root, this is where the plus-minus comes from ...
A: Per the comments (and making this CW to avoid reputation gain for others' work), the issue is how we interpret the symbol "$\sqrt\cdot$" - is it the function outputting the unique nonnegative square root of a nonnegative input, or is it describing the set of all square roots of the input?
The standard meaning is the former, so e.g. $\sqrt{4}=2$. Regardless, the notation "$\pm$" only adds clarity to the situation.

Incidentally, thinking ahead to the complex numbers, note that while $\{x: x^2=a\}$ always makes sense there are serious problems with trying to produce a "square root function" in the context of $\mathbb{C}$, and more generally with exponentiation in the complex numbers in general. There are lots of posts relating to this on the site - see e.g. here or here.
