Solve the quadratic equation by completing the square. (Enter your answers as a comma-separated list.) \( x=\square \) \( +\quad+\quad x^{2}+4 x-1=0 \)

We start with the equation \[ x^2 + 4x - 1 = 0. \] 1. Move the constant term to the right-hand side: \[ x^2 + 4x = 1. \] 2. To complete the square, find half of the coefficient of \(x\), which is \(\frac{4}{2} = 2\). Square it to get \(2^2 = 4\). Add \(4\) to both sides: \[ x^2 + 4x + 4 = 1 + 4. \] 3. The left-hand side is now a perfect square trinomial: \[ (x + 2)^2 = 5. \] 4. Take the square root of both sides, remembering to include both the positive and negative roots: \[ x + 2 = \pm \sqrt{5}. \] 5. Solve for \(x\) by subtracting \(2\) from both sides: \[ x = -2 \pm \sqrt{5}. \] Thus, the solutions are: \[ \sqrt{5}-2,\quad -\sqrt{5}-2. \]