\(C=\frac{2x^2-3x-9}{x^2-9}=0\)
\(\Leftrightarrow2x^2-3x-9=0\)
Ta có : \(\Delta=\left(-3\right)^2-4.2.\left(-9\right)=81>0\)
\(\Rightarrow x1=\frac{3-\sqrt{81}}{2.2}=\frac{3-9}{4}=-\frac{3}{2}\)
\(\Rightarrow x2=\frac{3+\sqrt{81}}{2.2}=\frac{3+9}{4}=3\)