Nothing about the standard quadratic formula is really intuitive. Sure, you can derive it by completing the square, but that gets complicated, and isn't really an accessible proof on the level of those learning to solve quadratics for the first time. However, Loh's method builds on an understanding of both factoring and graphing.
For example, $x^2 – 10x + 21$ factors as $(x-3)(x-7)$ and therefore has the solutions $3$ and $7$. Notice that $B=-10=-(3+7)$ and $C=21=(3)(7)$. Therefore $-B$ is the sum of the solutions, and $C$ is the product of the solutions. Both of these facts will be needed.
Now consider the graph of $y=x^2 – 10x + 21$ shown below. To use Loh's method, we'll need two other variables: $m$ and $d$. Where $(m, 0)$ is the midpoint of the zeros, $m$ is the average of the solutions. Then $d$ is the distance each zero is from the midpoint. Therefore, we can represent the solutions as $m-d$ and $m+d$ or as just $m \pm d$. If we could calculate $m$ and $d$ simply from $B$ and $C$, we'd have an easy way to solve a quadratic. And we can!

Let's get to Loh's method. We'll begin by assuming we have a quadratic of the form $Ax^2+Bx+C=0$ where $A=1$. We've already established that $-B$ is the sum of our solutions. Since the mean of the solutions is their sum divided by 2, $m=\frac{-B}{2}$. Also recall that $C$ is the product of the solutions. Therefore, $C=(m-d)(m+d)=m^2-d^2$. If we rearrange this as $d^2=m^2-C$, we have an easy way to find $d$ from $m$ and $C$. We can then write our solutions as $m \pm d$.
Here's how it works out with $y=x^2 – 10x + 21$.
$m=\frac{-B}{2}=\frac{10}{2}=5$
$d^2=m^2-C=(5)^2-21=4$
Therefore, $d=\pm \sqrt 4=\pm 2$
Since $m\pm d=5\pm 2$, the solutions are 3 and 7.
That's Loh's method! Again, it's far more accessible to students just learning how to solve quadratics.
I will admit, no one talks much about the case where $A\neq 1$. Sure, 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 $3x^2 + 3x + 1 = 0$ 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. Plus, the connection made from here to the actual quadratic formula isn't nearly as intuitive and accessible as the rest of Loh's method. Before I posted this answer, I posted a related question and answer here that I think is better for when $A\neq 1$.