How can I simplify Po-Shen Loh's method of solving quadratics? My question is, is there a way to eliminate the need for fraction arithmetic and rationalizing denominators necessitated by Po-Shen Loh's alternative to the quadratic formula while still being able to provide a student to whom you're teaching the method an intuitive understanding of what is going on?
A few months back, Po-Shen Loh published an article providing an alternative to the quadratic formula. His method applies the properties of quadratics described in this answer.
For Loh's method to be used, the quadratic must be of the form $Ax^2+Bx+C=0$ where $A=1$. Certainly, if $A\neq 1$ you can divide through by $A$ and not affect the roots, but it means fractions, accompanying fraction arithmetic, and the possible need to rationalize denominators--all of which is not necessary if just using the commonly memorized quadratic formula. Consider just trying to solve $3x^2 + 3x + 1 = 0$ using Loh's method and you'll see what I mean. You end up with fractions all the way through with denominators of 2, 3, 4, 6, and 12 at some point in the process.
 A: Yes, there is! We'll just need to perform a dilation on the function by replacing $y$ with $\frac{y}{A}$ and $x$ with $\frac{x}{A}$. The result is a larger scaled version of the parabola with $A$ being the scale factor. You'll just need to divide your answer of $m\pm d$ by $A$. 
Notice what this dilation does to the function.  

$\frac{y}{A}=A(\frac{x}{A})^2+B(\frac{x}{A})+C$
$y=x^2+Bx+AC$

Now, the new $A$ is $1$, but neither the new $B$ nor the new $C$ are fractions making this quadratic SO much easier to solve! And, once you understand why, you can save time by simply replacing $A$ with $1$, and $C$ with $AC$ rather than going through the whole process. 
Furthermore, since $m=\frac{-B}{2}$, you might consider manipulating the function to avoid fractions caused by a $B$ that's not divisible by 2. To do so, first multiply through by $2$. This will not change the solutions since $0=2(Ax^2+Bx+C)$ has the same zeros as $0=Ax^2+Bx+C$. Next, dilate by a scale factor of $2A$ to eliminate the coefficient of $x^2$. For an original equation of $0=Ax^2+Bx+C$, the resulting equation would be $0=x^2+2Bx+4AC$--an even easier quadratic to solve! And all you'd need to do is scale back the solutions by $2A$ at the end!  
If something about this seems familiar, it should! Using this equation now DIRECTLY ties to the quadratic formula!  

$m=\frac{-2B}{2}=-B$
$d^2=m^2-C=B^2-4AC$
$d=\sqrt{B^2-4AC}$
$m\pm d=-B\pm \sqrt{B^2-4AC}$
  Scaled back by $2A$, our solutions are $\frac{-B\pm \sqrt{B^2-4AC}}{2A}$.  

As cool as it would be to stop right here, I think it'd be good to end with this being applied to an actual quadratic. We'll apply it to $3x^2 + 3x + 1 = 0$ taken from the end of the question. Again, this really is no different than using the quadratic formula, only its now intuitive--we know what all the pieces mean.


*

*Multiply through by $B$, and dilate using $2A$ as the scale factor by writing a new equation of the form $0=x^2+2Bx+4AC$.  


$0=x^2+6x+12$


*Find $m$ and $d$ for this new quadratic.  


$m=\frac{-6}{2}=-3$
$d^2=(-3)^2-12=-3$
$d=\sqrt{-3}=i\sqrt{3}$
$\frac{m\pm d}{2A}=\frac{-3\pm i\sqrt{3}}{6}$

