$\begin{align}  & \frac{6x-10}{{{x}^{2}}-2x-3}=\frac{6x-10}{(x-3)(x+1)} \\ & \frac{6x-10}{(x-3)(x+1)}=\frac{A}{(x-3)}+\frac{B}{x+1} \\ & 6x-10=A(x+1)+B(x-3) \\ & \text{Set }x=3 \\ & 6(3)-10=A(3+1) \\ & 8=4A \\ & A=\frac{8}{4}=2 \\ & \text{Set }x=-1 \\ & 6(-1)-10=B(-1-3) \\ & -16=-4B \\ & B=4 \\ & \text{The partial fraction becomes }\frac{2}{x-3}+\frac{4}{x-1} \\\end{align}$