Proving Quadratic Formula

$b/2a$ does NOT become $b^2/4a^2$. All that happens in the third row is that $b^2/4a^2$ is added to both sides of the equation.

The bit about taking half of the $x$ term and squaring it is just a means of working out WHAT to add. This is often called "completing the square" - adding a constant term to an expression to turn it into a perfect square, so that one may later take its square root.

Remember how to complete the square: $$Ax^2+Bx=A\left(x+\frac{B}{2A}\right)^2-\frac{B^2}{4A^2}$$

So now

$$ax^2+bx+c=0 ---- \text{complete square}$$ $$a\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a}=-c$$ $$\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}$$ $$x_{1,2}+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}$$ $$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

$$ax^2+bx+c=0 - \text{divide by $a$ because $a\neq 0$ }$$ 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 $$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$