<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] Every quadratic can have its square completed,  which is in fact where the quadratic formula comes from. 
The method I use is to factor out the leading coefficient,
$$
\begin{align}
\color{red}{a} x ^2 + bx +c &= \color{red}{a}\left( x^2 + \frac{b}{\color{red}{a}} x \right) + c \\
&= a\left( \left(x+\frac{b}{2a}\right)^2 - \frac{b^2}{4a^2} \right) + c \\
&= a\left(x+\frac{b}{2a}\right)^2 + c-\frac{b^2}{4a},
\end{align}
$$
if you solve for when this is equal to zero you will get the quadratic formula. 
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