<s>[INST] Proving Quadratic Formula purplemath.com explains the quadratic formula. I don't understand the third row in the "Derive the Quadratic Formula by solving $ax^2 + bx + c = 0$." section. How does $\dfrac{b}{2a}$ become $\dfrac{b^2}{4a^2}$?
 [/INST] $$ax^2+bx+c=0 - \text{divide by $a$ because $a\neq 0$ }$$ we get
$$x^2+\frac{b}{a}x+\frac{c}{a}=0$$
$$x^2+2x\frac{b}{2a}+\frac{c}{a}=0$$
$$x^2+2x\frac{b}{2a}+\frac{b^2}{4a^2}-\frac{b^2}{4a^2}+\frac{c}{a}=0$$
$$x^2+2x\frac{b}{2a}+\frac{b^2}{4a^2}=\frac{b^2}{4a^2}-\frac{c}{a}$$
$$x^2+2x\frac{b}{2a}+\frac{b^2}{4a^2}=\frac{b^2-4ac}{4a^2}$$
$$x^2+2x\frac{b}{2a}+\left(\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}$$
if in LHS we use $x=A$ and $\frac{b}{2a}=B$ then we have
$$A^2+2AB+B^2=(A+B)^2$$ or
$$\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}$$
we have two values of square roote
$$x_1+\frac{b}{2a}=+\sqrt{\frac{b^2-4ac}{4a^2}}$$and
$$x_2+\frac{b}{2a}=-\sqrt{\frac{b^2-4ac}{4a^2}}$$ or
$$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
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