Po-Shen Loh's new way of solving quadratic equation Quadratic equation, $ax^2+bx+c=0$ and its solution is quadratic equation, $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ 
Now setting $a=1$ then we have $x^2+bx+c=0$
$$x=\frac{-b\pm \sqrt{b^2-4c}}{2}$$ rewrite as 
$$x=-\frac{b}{2}\pm \sqrt{\left(\frac{b}{2}\right)^2-c}$$ 
In this new video Dr. Loh claims to discover a new way of solving the quadratic equation! How? It is the same as the above formula, by using the quadratic formula, the only thing I see different, is he rewrite it in the above form!
Can someone please explain to me how this is a new way?  

Suppose that $A$ is the arithmetic mean of the roots of the quadratic $P$, and $R$ is the geometric mean [where $P(x) = ax^2+bx+c = a(x-r_1)(x-r_2)$], then we have this equation:
$$\begin{align}
P(x) &= 0\\
ax^2+bx+c &= 0\\
x^2+\frac{b}{a}x+\frac{c}{a} &= 0\\
x^2+\beta x+\gamma &= 0\\
\end{align}$$
But we can also look at it from another point of view.
$$\begin{align}
a(x-r_1)(x-r_2) &= 0\\
(x-r_1)(x-r_2) &= 0\\
x^2 -(r_1+r_2)x+r_1r_2 &= 0\\
x^2 -2\bigg(\frac{r_1+r_2}{2}\bigg)x+\sqrt{r_1r_2}^2 &= 0\\
x^2 -2Ax+R^2 &= 0\\
\end{align}$$
which can be solved via completing the square.
$$\begin{align}
x^2-2Ax+R^2 &= 0\\
(x^2-2Ax+A^2)+(R^2-A^2) &= 0\\
(x-A)^2+(R^2-A^2) &= 0\\
(x-A)^2-(A+R)(A-R) &= 0\\
\dots\\
\end{align}$$
But isn't that just Dr. Loh's method?
