smc

By , we have $2x^2-x=3$. Subtracting $3$ from both sides, giving $2x^2-x-3=0$. This factors to $(2x-3)(x+1)=0$. Thus, $x=\frac{3}{2},-1$, so the equation is $\boxed{\text{Is satisified for only 2 values of }x}$. Note: Alternatively, one may note that the equation is quadratic with a nonzero discriminant, so it will be satisfied for exactly two values of $x$.