<s>[INST] Need proofread for deriving quadratic equation formula How to Solve quadratic equation $$ax^{2}+bx+c=0$$ such as $$a \neq 0$$
Divide by a both side from the equation
such as
$$\frac{a}{a}x^2 + \frac{b}{a}x + \frac{c}{a} = 0$$
$$\Rightarrow x^{2}+\frac{b}{a}x + \frac{c}{a} = 0$$
$$\Rightarrow x^{2} + \frac{b}{a}x + (\frac{b}{2a})^2 - (\frac{b}{2a})^2 + \frac{c}{a}=0$$
$$\Rightarrow (x + \frac{b}{2a})^2 - (\frac{b}{2a})^{2} + \frac{c}{a} = 0$$
$$\Rightarrow (x + \frac{b}{2a})^2 = (\frac{b}{2a})^2 - \frac{c}{a}$$
$$\Rightarrow (x + \frac{b}{2a})^2 = \frac{b^2}{4a^{2}} - \frac{c}{a}$$
$$\Rightarrow (x + \frac{b}{2a})^2 = \frac{b^{2}}{4a^{2}} - \frac{4ac}{4a^{2}}$$
$$\Rightarrow (x + \frac{b}{2a})^2 = \frac{b^{2} - 4ac}{4a^{2}}$$
$$\Rightarrow (x + \frac{b}{2a}) =\pm \sqrt{\frac{b^{2} - 4ac}{4a^{2}}}$$
$$\Rightarrow x = -\frac{b}{2a} \pm \sqrt{\frac{b^{2} - 4ac}{4a^{2}}}$$
$$\Rightarrow x = \frac{-b\pm\sqrt{b^{2} - 4ac}}{2a}$$
 [/INST] You have an error on your second-to-last line.
$$x + \frac{b}{2a} = \pm \sqrt{\frac{b^2 - 4ac}{4a^2}}$$
is correct, but when you move the $\frac{b}{2a}$ over, you forgot to change its sign. Correctly, this becomes
$$x + \frac{b}{2a} - \frac{b}{2a} = -\frac{b}{2a} \pm \sqrt{\frac{b^2 - 4ac}{4a^2}}$$
$$ \implies x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$.
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