If  $\frac{8x^2-12}{x(x^2+2x-6)}=\frac{A}{x}+\frac{Bx+C}{x^2+2x-6}$ then  $(A,C)=$

$8x^2-12=A(x^2+2x-6)+(Bx+C)x$

Put  $x=0\Rightarrow -12=-6A\Rightarrow A=2$

$\left.\begin{array}{r}\begin{array}{l}\text{Compare\hspace{0.17em}coefficent\hspace{0.17em}}{x}^2\text{then\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}A+B=8\\ \end{array}\\ \text{Compare\hspace{0.17em}eoefficent\hspace{0.17em}}x\text{\hspace{0.17em}then\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2A+C=0\end{array}\right\}\begin{array}{c}\text{B}=6\\ \\ \text{C}=-4\end{array}$

$\therefore (A,C)=(2,-4)$