Find the real solutions, if any, for the following equation. Use any method. \( x^{2}+x=1 \) Select the correct choice below and, if necessary, fill in the answer box to complete your answer. A. The solution set is \( \} \). Use a comma to separate answers as needed.) (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression. B. The solution is not a real number.

To solve the equation \( x^2 + x = 1 \), we first rearrange it into standard form: \[ x^2 + x - 1 = 0. \] Now, we can apply the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 1, \) and \( c = -1 \). Calculating the discriminant: \[ b^2 - 4ac = 1^2 - 4(1)(-1) = 1 + 4 = 5. \] Since the discriminant is positive, there are two real solutions. Plugging into the quadratic formula: \[ x = \frac{-1 \pm \sqrt{5}}{2}. \] Thus, the solution set is \( \left\{ \frac{-1 + \sqrt{5}}{2}, \frac{-1 - \sqrt{5}}{2} \right\} \). So the final answer is: A. The solution set is \( \left\{ \frac{-1 + \sqrt{5}}{2}, \frac{-1 - \sqrt{5}}{2} \right\} \). --- Did you know that the quadratic formula has its roots in ancient civilizations? The Babylonians were solving quadratic equations as early as 2000 BC! They had geometric interpretations for these solutions long before the formula was documented. Another fun fact is that every quadratic equation can be graphed as a parabola. For our equation \( x^2 + x - 1 = 0 \), the parabola opens upwards since the coefficient of \( x^2 \) is positive. This means there will be two points where the parabola crosses the x-axis, which corresponds to our two real solutions!