Correct. By subtracting the second equation from the first equation, you find that $\displaystyle{0}={2}{{x}}^{{2}}\usepackage{cmbright}\Huge\$ , and therefore $\displaystyle{x}={0}\usepackage{cmbright}\Huge\$ . Substituting $\displaystyle{x}={0}\usepackage{cmbright}\Huge\$ into the first equation, you get $\displaystyle{y}={{\left({0}\right)}}^{{2}}+{2}{\left({0}\right)}+{3}={3}\usepackage{cmbright}\Huge\$ , so the solutions is $\displaystyle{\left({0},{3}\right)}\usepackage{cmbright}\Huge\$ .