smc

If the system of equations has two identical solutions, then only one $(x,y)$ pair will satisfy both equations. Substitute $y$ in one equation into another equation. $$x^2=3x+k$$ $$x^2-3x=k$$ Complete the square to get $$x^2-3x+\frac{9}{4}=k+\frac{9}{4}$$ $$(x-\frac{3}{2}{)}^2=k+\frac{9}{4}$$ In order for the equation to have one solution, the right side must be $0$, so $k=-\frac{9}{4}$, which is answer choice $\boxed{-\frac{9}{4}}$.