<s>[INST] I think I can complete the square of any quadratic, is it true? (Any reason to ever use Quad. Formula?) I was taught that you could only complete the square of a quadratic if the coefficient on the $x^2$ term is 1. 
However, playing a little bit with other quadratics, I've found that it's just not true. Based on the CTS algorithm, you just need to divide the coefficient of the $x$ term by twice the square root of the coefficient of the $x^2$ term.
So, if you have $ax^2 + bx + c$, your perfect square would be $(\sqrt{a}x + \frac{b}{2 \sqrt{a}})^2$
If $a$ is not a perfect square it could get nasty, but then you can just square the whole quadratic and go from there.
For example: 
In the equation $5x^2 + 6x + 5 = 0$, we could do:
$25x^2 + 30x + 25 = 0$
$(5x+3)^2 = -16$
$5x+3 = \pm4i$
$x = \pm \frac{4i}{5} - \frac{3}{5}$ 
My questions are:
-Is this correct?
-Is there ever an advantage to using the quadratic formula?
-Are there quadratics that are unsolvable this way? 
 [/INST] If $a$ is not a “perfect square” there's no problem either. If your equation is
$$
ax^2+bx+c=0
$$
then it's equivalent to
$$
4a^2x^2+4abx+4ac=0
$$
and completing the square is more evident:
$$
4a^2x^2+4abx+b^2-b^2+4ac=0
$$
or
$$
(2ax+b)^2-(b^2-4ac)=0
$$
If you just want to factor the polynomial $ax^2+bx+c$ (with $a\ne0$, of course, just do the same:
$$
ax^2+bx+c=\frac{1}{4a}(4a^2x^2+4abx+4ac)=
\frac{1}{4a}\bigl((2ax+b)^2-(b^2-4ac)\bigr)
$$
If $b^2-4ac<0$ there's nothing else to do, because the polynomial is irreducible over the reals; if $b^2-4ac=0$ it is
$$
\frac{1}{4a}(2ax+b)^2
$$
and, if $b^2-4ac>0$ you get
$$
ax^2+bx+c=\frac{1}{4a}(x+2a-\sqrt{b^2-4ac})(x+2a+\sqrt{b^2-4ac})
$$
</s>