<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] For $b^2-4ac\geq0$ and $a=1$ they are the same:
$$\frac{-b+\sqrt{b^2-4ac}}{2a}=\frac{-\frac{b}{2}+\sqrt{\frac{b^2-4ac}{4}}}{a}=\frac{-\frac{b}{2}+\sqrt{\left(\frac{b}{2}\right)^2-ac}}{a}=-\frac{b}{2}+\sqrt{\left(\frac{b}{2}\right)^2-c}$$ and 
$$\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{-\frac{b}{2}-\sqrt{\frac{b^2-4ac}{4}}}{a}=\frac{-\frac{b}{2}-\sqrt{\left(\frac{b}{2}\right)^2-ac}}{a}=-\frac{b}{2}-\sqrt{\left(\frac{b}{2}\right)^2-c}.$$
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