<s>[INST] Roots of quadratic equation are given by $b \pm \sqrt{b^2 - c}$ I was reading slides about the cancellation error in quadratic equations and it's written:

The roots of the quadratic equation:
$$x^2 - 2bx + c = 0$$
with $b^2 > c$ are given by $b \pm \sqrt{b^2 - c}$.

Fact that let me perplexed, since I always thought that the roots can be found using the following formula:
$$\frac{b \pm \sqrt{b^2 - 4ac}}{2a}$$
What the relation between one and the other?
 [/INST] If we write the quadratic like this $$ax^2+bx+c=0,$$ then the quadratic formula is right as you have written it; the roots are: $$\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$ But the quadratic $$x^2-2bx+c=0$$ has replaced $b$ with $-2b$ and $a$ with $1$. This makes the quadratic $$\frac{2b\pm\sqrt{(-2b)^2-4(1)c}}{2(1)}=\frac{2b\pm2\sqrt{b^2-c}}{2}=b\pm\sqrt{b^2-c}.$$
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