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

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  • $\begingroup$ 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. $\endgroup$ – Tobias Kildetoft Aug 7 '13 at 8:40
  • $\begingroup$ You might wish to take a look at this answer. $\endgroup$ – Ben Aug 7 '13 at 9:27
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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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