1/s+a + 1/s+b = 1/c
solve for s
$\displaystyle \frac{1}{s + a} + \frac{1}{s + b} = \frac{1}{c}$
$\displaystyle \frac{s + b}{(s + a)(s + b)} + \frac{s + a}{(s + a)(s + b)} = \frac{1}{c}$
$\displaystyle \frac{s + b + s + a}{(s + a)(s + b)} = \frac{1}{c}$
$\displaystyle \frac{2s + a + b}{s^2 + (a + b)s + ab} = \frac{1}{c}$
$\displaystyle c(2s + a + b) = s^2 + (a + b)s + ab$
$\displaystyle 2cs + ac + bc = s^2 + (a + b)s + ab$
$\displaystyle 0 = s^2 + (a + b - 2c)s + ab - ac - bc$.
Now solve using the Quadratic Formula...
No, the OP is asking what is a, b, and c in the quadratic formula. So in the equation
$\displaystyle s^2 + (a + b - 2c)s + ab - ac - bc = 0$
"a" is 1,
"b" is a + b - 2c, and
"c" is ab - ac - bc.
So the setup would be
$\displaystyle s = \frac{-(a + b - 2c) \pm \sqrt{(a + b - 2c)^2 - 4(1)(ab - ac - bc)}}{2(1)}$
...