Answer
a) $11.8\mu C$ and $78.2\mu C$ or $88.8\mu C$ and $1.2\mu C$
b) $-91.2\mu C$ and $1.2\mu C$
Work Step by Step
Lets call the first charge $Q_{1}$ and the second charge $Q_{2}$. The sum of these two charges are given as $90.0\mu C=9\times10^{-5}C$ and the distance between them $28.0cm=0.28m$.
Applying Coulomb's law,
$F=\frac{kQ_{1}Q_{2}}{r^{2}}=\frac{(9.0\times10^{9}\frac{N\times m^{2}}{C})Q_{1}Q_{2}}{0.28m^{2}}=12.0N$
Rearranging and solving, we get two equations and two unknowns.
$Q_{1}\times Q_{2}=1.05\times10^{-10} C$
$Q_{1}+Q_{2}=9\times10^{-5}C$
$Q_{1}=9\times10^{-5}C-Q_{2}$
Substituting this into the first equation and simplifying, we get the following quadratic equation.
$Q_{2}^{2}-(9\times10^{-5})Q_{2}+(1.05\times10^{-10})=0$
The two solutions are: $11.8\mu C$ and $78.2\mu C$
or
$88.8\mu C$ and $1.2\mu C$
However, if the two charges were attractive, the difference of the two charges will be equal to $90.0\mu C$ and the quadratic equation will be
$Q_{2}^{2}+(9\times10^{-5})Q_{2}-(1.05\times10^{-10})=0$
The two solutions are: $-91.2\mu C$ and $1.2\mu C$
