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? 

It seems like your questions have been answered, so this response is more of an addition. Completing the square is a useful rearrangement in and of itself - beyond the usage for solving quadratics. You may be aware of its application in placing the equation of a conic section in standard form.
For example, consider
$$x^2 + 2x + 2y^2 + y + 3 = 6$$

This is an ellipse, but our equation does not tell us much information. Rearranging into standard form will reveal quite a bit of info. Standard form for a horizontal ellipse is given as
$$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$$
where $(h,k)$ is the center and $a$ and $b$ represent the semi-major and semi-minor axes respectively (the longest and shortest "radius" of this "eccentric circle").
In order to place our ellipse into standard form, we complete the square twice:
$$x^2 + 2x + 2y^2 + 4y + 3 = 6$$
$$\left(x^2 + 2x \color{red}{+1}\right) + 2\left(y^2 + 4y \color{red}{+4}\right)+ 3 = 6  \color{red}{+1} \color{red}{+8} $$
$$(x+1)^2+ 2(y+2)^2+ 3 = 15 $$
$$(x+1)^2+ 2(y+2)^2 = 12 $$
$$\frac{(x+1)^2}{12}+ \frac{(y+2)^2}{6} = 1 $$
We now see that we have an ellipse centered at $(h,k) = (-1,-2)$ with semi-major axis $a = \sqrt{12}$ and semi-minor axis $b=\sqrt6$. This matches the graph.
