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?
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 $$\frac{2b\pm\sqrt{(-2b)^2-4(1)c}}{2(1)}=\frac{2b\pm2\sqrt{b^2-c}}{2}=b\pm\sqrt{b^2-c}.$$
One formula solves the equation $x^2-2bx+c=0$, the other solves $ax^2+bx+c=0$.
Use the usual formula for your equation $\;x^2-2bx+c=0\;$ :
$$x_{1,2}=\frac{2b\pm\sqrt{4b^2-4c}}2=\frac{2b\pm2\sqrt{b^2-c}}2=b\pm\sqrt{b^2-c}$$
They use the so-called reduced formulae, when the coefficient of $x$ has the form $b=2b'$. In this case the discriminant is $\;\Delta=4b'^2-4ac=4\Delta'\;$ (reduced discriminat), and the formulae become $$x=\frac{-b\pm\sqrt{\Delta}}{2a}=\frac{-2b'\pm2\sqrt{\Delta'}}{2a}=\color{red}{\frac{-b'\pm\sqrt{\Delta'}}{a}}$$ Add the fact that here we have a monic polynomial ($a=1$) and you gete the mentioned formula.
With the usual formula
$$x^2-2bx+c=0$$ has the roots
$$\frac{-(-2b)\pm\sqrt{(-2b)^2-4\cdot1\cdot c}}{2\cdot1}=b\pm\sqrt{b^2-c}.$$
Never mind, they can be found by that formula only for this quadratic equation $$x^2 - 2bx + c = 0$$
Indeed if we use the usual formula:
$$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
to the equation above we obtain:
$$\frac{-(-2b) \pm \sqrt{(-2b)^2 - 4*1*c}}{2*1}$$
$$\frac{2b \pm \sqrt{4b^2 - 4c}}{2}$$
$$\frac{2b \pm \sqrt{4(b^2 - c)}}{2}$$
$$\frac{2b \pm 2\sqrt{b^2 - c}}{2}$$
$$\frac{2(b \pm \sqrt{b^2 - c})}{2}$$
$$b \pm \sqrt{b^2 - c}$$
