Answer
{$2-\sqrt 5,2+\sqrt 5$}
Work Step by Step
Step 1: Comparing $n^{2}-4n-1=0$ to the standard form of a quadratic equation $ax^{2}+bx+c=0$, we obtain:
$a=1$, $b=-4$ and $c=-1$
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:
$x=\frac{-(-4) \pm \sqrt {(-4)^{2}-4(1)(-1)}}{2(1)}$
Step 4: $x=\frac{4 \pm \sqrt {16+4}}{2}$
Step 5: $x=\frac{4 \pm \sqrt {20}}{2}$
Step 6: $x=\frac{4 \pm \sqrt {4\times5}}{2}$
Step 7: $x=\frac{4 \pm \sqrt {2^{2}\times5}}{2}$
Step 8: $x=\frac{4 \pm 2\sqrt {5}}{2}$
Step 9: $x=2\pm\sqrt 5$
Step 10: $x=2+\sqrt 5$ or $x=2-\sqrt 5$
Step 11: Therefore, the solution set is {$2-\sqrt 5,2+\sqrt 5$}.
