Condition \(x\ne-2\)
We have :
\(\dfrac{x^2-4}{x+2}=\dfrac{3x}{2}\)
\(\Leftrightarrow\dfrac{\left(x-2\right)\left(x+2\right)}{x+2}=\dfrac{3x}{2}\)
\(\Leftrightarrow x-2=\dfrac{3x}{2}\)
\(\Leftrightarrow2\left(x-2\right)=3x\)
\(\Leftrightarrow2x-4=3x\)
\(\Leftrightarrow x=-4\)
So the value of x is : \(-4\)