What is the solution set for this quadratic equation? x^2-2x-4=0 A 1-sqrt (10),1+sqrt (10)) B ) there is no real solution 2,4 D 1-sqrt (5),1+sqrt (5))

To solve the quadratic equation \(x^2 - 2x - 4 = 0\), we can use the quadratic formula:<br /><br />\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]<br /><br />Here, \(a = 1\), \(b = -2\), and \(c = -4\). Plugging these values into the quadratic formula gives:<br /><br />\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-4)}}{2(1)} \]<br />\[ x = \frac \pm \sqrt{4 + 16}}{2} \]<br />\[ x = \frac{2 \pm \sqrt{20}}{2} \]<br />\[ x = \frac{2 \pm 2\sqrt{5}}{2} \]<br />\[ x = 1 \pm \sqrt{5} \]<br /><br />So, the solution set is:<br /><br />\[ \{ 1 - \sqrt{5}, 1 + \sqrt{5} \} \]<br /><br />Therefore, the correct answer is:<br /><br />D. \(\{ 1 - \sqrt{5}, 1 + \sqrt{5} \}\)