<s>[INST] Quadratic formula $x = \frac{- (b +\sqrt{b^2- 4ac})}{ \pm2a}$ In the proof of the quadratic formula
$$x = \frac{- b +\sqrt{b^2- 4ac}}{2a}$$
shouldn't there be $\pm 2a$ instead of $2a$, since both can be  the square root of $4a^2$?
 [/INST] I will make a guess that the proof you were looking at has the following equation in one of its steps:
$$ \left(x+\frac{b}{2a} \right)^2 = \frac{b^2-4ac}{4a^2}. $$
This tells us that $x+\dfrac{b}{2a}$ is one of the square roots
of $\dfrac{b^2-4ac}{4a^2}.$
But it could be either the positive or negative square root.
The square roots of $\dfrac{b^2-4ac}{4a^2}$ are
$\dfrac{\sqrt{b^2 - 4ac}}{2a}$ and $-\dfrac{\sqrt{b^2 - 4ac}}{2a}.$
You can verify this by squaring each one:
\begin{align}
\left(\frac{\sqrt{b^2 - 4ac}}{2a}\right)^2 &= \frac{b^2-4ac}{4a^2}, \\
\left(-\frac{\sqrt{b^2 - 4ac}}{2a}\right)^2 &= \frac{b^2-4ac}{4a^2}. \\
\end{align}
Notice that $-\dfrac{\sqrt{b^2 - 4ac}}{2a}$ is just
$\dfrac{\sqrt{b^2 - 4ac}}{2a}$ with its sign flipped.
That is, one of these is a positive square root and one is a negative square root.
Which one is the positive root and which is the negative root depends on the sign of $a,$ but no matter which sign $a$ has we still have both roots, the positive and the negative.
Therefore we have found that the following statement is true:
$$
 \text{$x + \frac{b}{2a}$ is either
 $\frac{\sqrt{b^2 - 4ac}}{2a}$ or $-\frac{\sqrt{b^2 - 4ac}}{2a}$}.
$$
Another way to write the same statement is
$$ x + \dfrac{b}{2a} = \pm\dfrac{\sqrt{b^2 - 4ac}}{2a}. $$
Now just subtract $\dfrac{b}{2a}$ from both sides and you have the usual quadratic formula.
At no point in any of this did we ever take a square root of $4a^2.$
Instead, we looked at the expression $\dfrac{b^2-4ac}{4a^2}$ and found its square roots
— both of them — and included both roots in the answer.
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