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**e^(i*pi)** You can use it because b^2 is a constant in terms of s (ie: there is no s term there)

Let a =p and b=q

s² + 2abs + b² = s² + 2pqs + q^2 = 0 (this will make the next step easier to understand),

In this equation we know that a=1, b = 2pq and c=q^2 (the coefficients of s^2, s and constant respectively)

$\displaystyle

s = \frac{-b\pm \sqrt{b^2-4ac}}{2a} = \frac{-2pq \pm \sqrt{4p^2q^2 - 4q^2}}{2}$

Looking at $\displaystyle \sqrt{4p^2q^2 - 4q^2}$ we can see that 4q² is a factor so we can remove that to give $\displaystyle \sqrt{4q^2} \sqrt{p^2-1} = 2q \sqrt{p^2-1} $

$\displaystyle

s = \frac{-b\pm \sqrt{b^2-4ac}}{2a} = \frac{-2pq \pm 2q\sqrt{p^2-1}}{2}$$\displaystyle = \frac{2q(-2 {\color{red}p}\pm \sqrt{p^2-1})}{2} = q(-2{\color{red}p}\pm \sqrt{p^2-1})$

If we reintroduce a and b for p and q we get a solution of

$\displaystyle s = b(-2{\color{red}a}\pm \sqrt{a^2-1})$

From there you put in your values of a and b