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 $$\frac{a}{a}x^2 + \frac{b}{a}x + \frac{c}{a} = 0$$ $$\Rightarrow x^{2}+\frac{b}{a}x + \frac{c}{a} = 0$$ $$\Rightarrow x^{2} + \frac{b}{a}x + (\frac{b}{2a})^2 - (\frac{b}{2a})^2 + \frac{c}{a}=0$$ $$\Rightarrow (x + \frac{b}{2a})^2 - (\frac{b}{2a})^{2} + \frac{c}{a} = 0$$ $$\Rightarrow (x + \frac{b}{2a})^2 = (\frac{b}{2a})^2 - \frac{c}{a}$$ $$\Rightarrow (x + \frac{b}{2a})^2 = \frac{b^2}{4a^{2}} - \frac{c}{a}$$ $$\Rightarrow (x + \frac{b}{2a})^2 = \frac{b^{2}}{4a^{2}} - \frac{4ac}{4a^{2}}$$ $$\Rightarrow (x + \frac{b}{2a})^2 = \frac{b^{2} - 4ac}{4a^{2}}$$ $$\Rightarrow (x + \frac{b}{2a}) =\pm \sqrt{\frac{b^{2} - 4ac}{4a^{2}}}$$ $$\Rightarrow x = -\frac{b}{2a} \pm \sqrt{\frac{b^{2} - 4ac}{4a^{2}}}$$ $$\Rightarrow x = \frac{-b\pm\sqrt{b^{2} - 4ac}}{2a}$$

• You are missing a minus on the $b$ near the end when you move it to the other side. Also, for this to be complete, you need to argue that the implication also go the other way. Commented Aug 7, 2013 at 8:40
• You might wish to take a look at this answer.
– Ben
Commented Aug 7, 2013 at 9:27

$$x + \frac{b}{2a} = \pm \sqrt{\frac{b^2 - 4ac}{4a^2}}$$
is correct, but when you move the $\frac{b}{2a}$ over, you forgot to change its sign. Correctly, this becomes
$$x + \frac{b}{2a} - \frac{b}{2a} = -\frac{b}{2a} \pm \sqrt{\frac{b^2 - 4ac}{4a^2}}$$
$$\implies x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$.