Answer
$s=\frac{2c-b-a\pm\sqrt{a^2+b^2+4c^2-2ab}}{2}$
Work Step by Step
$\frac{1}{s+a}+\frac{1}{s+b}=\frac{1}{c}$ (Multiply by $c(s+a)(s+b)$)
$c(s+b)+c(s+a)=(s+a)(s+b)$ (Simplify)
$2cs+(ac+bc)=s^2+(a+b)s+ab$
$s^2+(a+b-2c)s+(ab-ac-bc)=0$
Find the discriminant of the last equation:
$D=(a+b-2c)^2-4\cdot 1\cdot (ab-ac-bc)$
$D=a^2+b^2+4c^2+2ab-4ac-4bc-4ab+4ac+4bc$
$D=a^2+b^2+4c^2-2ab$
Using the quadratic formula,
$s=\frac{-(a+b-2c)\pm \sqrt{D}}{2\cdot 1}$
$s=\frac{2c-b-a\pm\sqrt{a^2+b^2+4c^2-2ab}}{2}$
