$\begin{array}{l}x^{2}+y^{2}=5 \\ y=-2 x\end{array}$

Step 1 ：The system of equations is: \(x^{2}+y^{2}=5\) and \(y=-2x\)

Step 2 ：Substitute equation \(y=-2x\) into equation \(x^{2}+y^{2}=5\): \(x^{2}+(-2x)^{2}=5\)

Step 3 ：Simplify: \(x^{2}+4x^{2}=5\)

Step 4 ：Combine like terms: \(5x^{2}=5\)

Step 5 ：Divide both sides by 5: \(x^{2}=1\)

Step 6 ：Take the square root of both sides: \(x=\pm1\)

Step 7 ：Substitute \(x=1\) into equation \(y=-2x\): \(y=-2(1)\), so \(y=-2\)

Step 8 ：Substitute \(x=-1\) into equation \(y=-2x\): \(y=-2(-1)\), so \(y=2\)

Step 9 ：So, the solutions to the system of equations are \((1,-2)\) and \((-1,2)\)

Step 10 ：Check these solutions: For \((1,-2)\): \((1)^{2}+(-2)^{2}=5\), which simplifies to \(1+4=5\), so \(5=5\)

Step 11 ：For \((-1,2)\): \((-1)^{2}+(2)^{2}=5\), which simplifies to \(1+4=5\), so \(5=5\)

Step 12 ：Both solutions are correct, so the final answer is \(\boxed{(1,-2)}\) and \(\boxed{(-1,2)}\)