If $\frac{3x+2}{\left(x+1\right)\left(2x^2+3\right)}=\frac{A}{x+1}+\frac{Bx+C}{2x^2+3}$ , then $A-B+C=$

Put $x=-1\Rightarrow -1=5A\Rightarrow A=\frac{-1}{5}$

Put $x=0\Rightarrow 2=3A+C\Rightarrow C=\frac{13}{5}$

Put $x=1\Rightarrow 5=5A+2B+2C\Rightarrow 5=-1+2B+\frac{26}{5}$

$\Rightarrow 2B=6-\frac{26}{5}\Rightarrow B=\frac{2}{5}$

$A-B+C=\frac{-1}{5}-\frac{2}{5}+\frac{13}{5}=2$