Root locus, where do they leave the line parallel to imaginary axis
First, maybe you should not use \$P\$ as a notation for the breakaway points, since your formula use it for the transfer function poles. The poles are the value for witch the denominator is 0. Let's call your breakaway points, the "points of interest", \$B_1\$ and \$B_2\$ (and \$B_3\$ for the real one).

\$G(s)\$ has 4 poles:


\$P_1=-4\$ for \$(s+4)\$
\$P_2=-6\$ for \$(s+6)\$
\$P_3=-5+j\sqrt{75}\$, \$P_4=-5+j\sqrt{75}\$, the roots of \$(s^2+10s+100)\$.


\$G(s)\$ has no zeroes, there is no \$N\$.

To get the breakaway points, you need to solve:
$$
\sum_{i=1}^4 \frac{1}{\sigma-P_i}=0
$$

That is also:
$$
\frac{1}{\sigma+4}+\frac{1}{\sigma+6}+\frac{1}{\sigma+5+j\sqrt{75}}+\frac{1}{\sigma+5-j\sqrt{75}}=0
$$
$$
\frac{2\sigma+10}{(\sigma+4)(\sigma+6)}+\frac{2\sigma+10}{\sigma^2+10\sigma+100}=0
$$
$$
(\sigma+5)\left[\frac{1}{(\sigma+4)(\sigma+6)}+\frac{1}{\sigma^2+10\sigma+100}\right]=0
$$
$$
(\sigma+5)\left[\frac{\sigma^2+10\sigma+62}{(\sigma+4)(\sigma+6)(\sigma^2+10\sigma+100)}\right]=0
$$
The solutions to that equation are \$\sigma_1=-5\$ from \$(\sigma+5)\$, and \$\sigma_2=-5+j\sqrt{37}\$, \$\sigma_3=-5-j\sqrt{37}\$ from \$(\sigma^2+10\sigma+62)\$. Note that \$\sqrt{37}\approx6.0828\approx6\$.

Here you go, the coordinates of \$B_1\$ are given by \$\sigma_2\$, the ones of \$B_2\$ are given by \$\sigma_3\$.