$\begin{array}{l}y=(2 x+1)^{2}-3 \\ y=-x+3\end{array}$

Step 1 ：Set the two equations equal to each other: \((2x + 1)^2 - 3 = -x + 3\)

Step 2 ：Solve for x: \(x = -\frac{5}{8} + \frac{\sqrt{105}}{8}\) and \(x = -\frac{\sqrt{105}}{8} - \frac{5}{8}\)

Step 3 ：Plug the x values back into either equation to find the corresponding y values: \(y = \frac{29}{8} - \frac{\sqrt{105}}{8}\) and \(y = \frac{\sqrt{105}}{8} + \frac{29}{8}\)

Step 4 ：\(\boxed{(-\frac{5}{8} + \frac{\sqrt{105}}{8}, \frac{29}{8} - \frac{\sqrt{105}}{8})}\) and \(\boxed{(-\frac{\sqrt{105}}{8} - \frac{5}{8}, \frac{\sqrt{105}}{8} + \frac{29}{8})}\) are the points of intersection