<s>[INST] 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?  
 [/INST] The teacher's conclusion at the end "guesswork has been replaced by a clever trick" implies that the main result of the proposed new method is the clever trick (change of unknowns $x_1$ and $x_2$) of solving the system of equations (which is the Vieta's theorem):
$$\begin{cases}x_1+x_2=-b\\ x_1x_2=c\end{cases} \stackrel{x_1=\frac{-b}{2}-t\\x_2=\frac{-b}{2}+t}\Rightarrow x_1x_2=\frac{b^2}{4}-t^2=c \Rightarrow t=\pm \sqrt{\frac{b^2}{4}-c} \Rightarrow \\
x_1=\frac{-b}{2}-\sqrt{\frac{b^2}{4}-c}=\frac{-b-\sqrt{b^2-4c}}{2}\\
x_2=\frac{-b}{2}+\sqrt{\frac{b^2}{4}-c}=\frac{-b+\sqrt{b^2-4c}}{2}$$
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