Answer
{$1-\sqrt 6,1+\sqrt 6$}
Work Step by Step
Step 1: Comparing $n^{2}-2n-5=0$ to the standard form of a quadratic equation $ax^{2}+bx+c=0$, we obtain:
$a=1$, $b=-2$ and $c=-5$
Step 2: The quadratic formula is:
$x=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$
Step 3: Substituting the values of a, b and c in the formula:
$x=\frac{-(-2) \pm \sqrt {(-2)^{2}-4(1)(-5)}}{2(1)}$
Step 4: $x=\frac{2 \pm \sqrt {4+20}}{2}$
Step 5: $x=\frac{2 \pm \sqrt {24}}{2}$
Step 6: $x=\frac{2 \pm \sqrt {4\times6}}{2}$
Step 7: $x=\frac{2 \pm \sqrt {2^{2}\times6}}{2}$
Step 8: $x=\frac{2 \pm 2\sqrt {6}}{2}$
Step 9: $x=1\pm\sqrt 6$
Step 10: $x=1+\sqrt 6$ or $x=1-\sqrt 6$
Step 11: Therefore, the solution set is {$1-\sqrt 6,1+\sqrt 6$}.
