So the proof goes like this:
$$f(x)=ax^2+bx+c$$
(1) "factor out the coefficient of $a$ from $x^2$ and $x$"
$$=a\left(x^2+\frac{b}{a}x\right)+c$$
(2) "add and subtract $\left(\frac{1}{2}\cdot\frac{b}{a}\right)^2$ to $x^2 + \frac{b}{a}x$"
$$=a\left(x^2+\frac{b}{a}x+\frac{b^2}{4a^2}-\frac{b^2}{4a^2}\right)+c$$
(3) "group the perfect square trinomial"
$$=a\left(x+\frac{b}{a}x+\frac{b^2}{4a^2}\right)+a\left(-\frac{b^2}{4a^2}\right)+c$$
(4) "factor and simplify"
$$a\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a}+c$$
(5) "obtain common denominator"
$$a\left(x+\frac{b}{2a}\right)^2+\frac{4ac-b^2}{4a}$$
I'm lost by two steps in the process, going from (3) to (4) what happens to the $a$ in front of $a\left(-\frac{b^2}{4a^2}\right)$ and from (4) to (5) the common factor of $-\frac{b^2}{4a}+c$ does not make sense to me.
At 3) we have $-a\frac{b^2}{4a^2}+c$
What happens to the $a$ in the first term? It cancels with one an $a$ in the denominator of the other factor.
$-\frac{b^2}{4a}+c$
Then we put everything over a common denominator of $4a$
$\frac{-b^2}{4a}+\frac {4ac}{4a}$
And add the two together
$\frac{-b^2 + 4ac}{4a}$
But starting with a negative sign looks wrong.
$\frac{4ac-b^2}{4a}$
Which is as far as this question goes. But then next thing you might want to know is for what values of $x$ does $f(x) = 0$?
These are the "roots"
$a(x+\frac {b}{2a})^2 + \frac {4ac - b^2}{4a} = 0\\ a(x+\frac {b}{2a})^2 = \frac {b^2 - 4ac}{4a}\\ (x+\frac {b}{2a})^2 = \frac {b^2 - 4ac}{4a^2}\\ x+\frac {b}{2a} = \pm\sqrt {\frac {b^2 - 4ac}{4a^2}}\\ x+\frac {b}{2a} = \pm\frac {\sqrt {b^2 - 4ac}}{2a}\\ x = \frac {-b \pm \sqrt {b^2 - 4ac}}{2a}$