<s>[INST] Help in getting the Quadratic Equation I'm starting a chapter on Functions and they had the steps shown to reach the p-q equation.
$$ x_{1,2} = -\frac{p}{2} \pm\sqrt{\left(\frac{p}{2}\right)^2 - q}$$ 
So I wanted to do the same with the Quadratic Equation. I'm using the base linear equation
$$ax+by+c = 0.$$
The solution I have so far is as follows:
$$x^2 + \frac{b}{a}x + \frac{c}{a}= 0$$
$$x^2 + \frac{b}{a}x = -\frac{c}{a}$$
$$x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 = -\frac{c}{a} + \left(\frac{b}{2a}\right)^2$$
$$\left(x + \frac{b}{2a}\right)^2 = \left(\frac{b}{2a}\right)^2 - \frac{c}{a}$$
$$\left(x + \frac{b}{2a}\right) = \pm\sqrt{\left(\frac{b}{2a}\right)^2 - \frac{c}{a}}$$
$$x = -\frac{b}{2a} \pm\sqrt{\left(\frac{b}{2a}\right)^2 - \frac{c}{a}}$$
My problem comes from trying to solve the insides of the square root:
$$\sqrt{(\frac{b}{2a})^2 - \frac{c}{a}} = \sqrt{\frac{b^2}{4a2} - \frac{c}{a}}$$
$$= \sqrt{\frac{b^2}{4a^2} - \frac{c}{a} \left(\frac{4a}{4a}\right)} = \sqrt{\frac{b^2 - 4ac}{4a^2}}$$
$$= \sqrt{\frac{b^2 - 4ac}{\left(2a\right)^2}}$$
Then:
$$x_{1,2} = \frac{-\left(\frac{b}{2a}\right) \pm\sqrt{b^2 -4ac}}{2a}$$
but there is still the problem of the -(b/2a) outside of the sqrt. 
What am I doing wrong? Also, Tex is awesome; is there a better way to do the 1,2 subscripts than _1,_2?
Answer:
Instead of 
$$x_{1,2} = \frac{-\left(\frac{b}{2a}\right) \pm\sqrt{b^2 -4ac}}{2a}.$$
The solution goes
$$\frac{-b}{2a}\pm \sqrt{\frac{b^{2}-4ac}{4a^{2}}}=\frac{-b}{2a}\pm \frac{\sqrt{b^{2}-4ac}}{2a}$$
$$= \frac{-b \pm\sqrt{b^2 - 4ac}}{2a}$$
 [/INST] You factored 2a out of the square root and put it in the denominator without factoring it out of -(b/2a).
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