$\begin{array}{l}y=x^{2}-2 x+1 \\ y=-x^{2}+3 x+4\end{array}$

Step 1 ：Given the two equations: \(y=x^{2}-2 x+1\) and \(y=-x^{2}+3 x+4\)

Step 2 ：To find the solution, we can set the two equations equal to each other, since they both equal to \(y\)

Step 3 ：So, \(x^{2}-2 x+1=-x^{2}+3 x+4\)

Step 4 ：Rearrange the equation, we get \(2x^{2}-5x-3=0\)

Step 5 ：This is a quadratic equation in the form of \(ax^{2}+bx+c=0\), where \(a=2\), \(b=-5\), and \(c=-3\)

Step 6 ：We can solve for \(x\) using the quadratic formula: \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\)

Step 7 ：Substitute \(a=2\), \(b=-5\), and \(c=-3\) into the formula, we get \(x=\frac{5\pm\sqrt{(-5)^{2}-4*2*(-3)}}{2*2}\)

Step 8 ：Simplify the equation, we get \(x=\frac{5\pm\sqrt{25+24}}{4}\)

Step 9 ：Further simplify the equation, we get \(x=\frac{5\pm\sqrt{49}}{4}\)

Step 10 ：So, \(x=\frac{5\pm7}{4}\)

Step 11 ：Therefore, the solutions are \(x=\frac{5+7}{4}=\frac{12}{4}=3\) and \(x=\frac{5-7}{4}=\frac{-2}{4}=-0.5\)

Step 12 ：Substitute \(x=3\) and \(x=-0.5\) into the first equation \(y=x^{2}-2 x+1\), we get \(y=3^{2}-2*3+1=4\) and \(y=(-0.5)^{2}-2*(-0.5)+1=2\)

Step 13 ：So, the solutions to the system of equations are \((3,4)\) and \((-0.5,2)\)

Step 14 ：Check the solutions by substituting them into the second equation \(y=-x^{2}+3 x+4\), we can confirm that they are correct

Step 15 ：So, the final solutions are \(\boxed{(3,4)}\) and \(\boxed{(-0.5,2)}\)