2. Find the solutions using the quadratic formula: \( x^{2}-5 x-1=0 \) A. \( x=\{7,-2\} \) B. \( x=\{2,-7\} \) C. \( x=\frac{-5+\sqrt{31}}{2}, \frac{-5-\sqrt{31}}{2} \) D. \( x=\frac{5+\sqrt{29}}{2}, \frac{5-\sqrt{29}}{2} \)

To solve the quadratic equation \( x^{2}-5x-1=0 \) using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we first identify \( a = 1 \), \( b = -5 \), and \( c = -1 \). Plug these values into the formula to find the discriminant \( b^2 - 4ac = (-5)^2 - 4(1)(-1) = 25 + 4 = 29 \). Now substituting back into the formula gives \( x = \frac{5 \pm \sqrt{29}}{2} \). Therefore, the correct answer is D: \( x=\frac{5+\sqrt{29}}{2}, \frac{5-\sqrt{29}}{2} \). A fun fact: Quadratic equations like this one pop up all the time in real life! Whether you're calculating the trajectory of a basketball shot, plotting the perfect parabola for a project, or optimizing profit in business, these equations have your back. They can help make sense of a variety of problems, translating seemingly chaotic situations into manageable mathematical models! Another fun tidbit: The quadratic formula, often attributed to ancient mathematicians, has a rich history dating back to Babylonian times. They had methods for solving quadratic equations long before algebra was formalized! Imagine how these early scholars viewed the world, laying the groundwork for problem-solving techniques that are now a fundamental part of mathematics education today.