$x^{2}+x-1=0$

Step 1 ：The given equation is a quadratic equation in the form \(ax^{2}+bx+c=0\), where \(a=1\), \(b=1\), and \(c=-1\).

Step 2 ：The solutions to a quadratic equation can be found using the quadratic formula: \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).

Step 3 ：Substituting the values of \(a\), \(b\), and \(c\) into the quadratic formula, we get two solutions: \(x1 = 0.6180339887498949\) and \(x2 = -1.618033988749895\).

Step 4 ：In exact form, the solutions are \(x=\frac{-1+\sqrt{5}}{2}\) and \(x=\frac{-1-\sqrt{5}}{2}\).

Step 5 ：Final Answer: The solutions to the equation \(x^{2}+x-1=0\) are \(x\approx0.618\) and \(x\approx-1.618\). Therefore, the solutions are \(\boxed{x=\frac{-1+\sqrt{5}}{2}, x=\frac{-1-\sqrt{5}}{2}}\).