Answer
The solution set is {$2-\sqrt 5, 2+\sqrt 5$}.
Work Step by Step
$x^2-4x-1=0$
$a=1, b=-4, c=-1$
$D=b^2-4ac=(-4)^2-4\cdot1(-1)=16+4=20>0$ The equation has two real solutions.
$x=\frac{-b\pm\sqrt {b^2-4ac}}{2a}=\frac{4\pm\sqrt 20}{2}$ Use quadratic formula.
$x=\frac{4\pm2\sqrt 5}{2}$ Simplify
$x=2\pm\sqrt 5$
$x=2+\sqrt 5$, $x=2-\sqrt 5$
The solution set is {$2-\sqrt 5, 2+\sqrt 5$}.
