Answer
$x = 2 \pm \sqrt3$
Work Step by Step
$Find$ $all$ $real$ $solutions$ $of$ $the$ $quadratic$ $equation:$
$0=x^2-4x+1$
Use the Quadratic Equation Formula: $x = \frac{-b \pm \sqrt {b^2-4ac}}{2a}$
$1x^2 -4x+1$
$a = 1, b = -4, c = 1$
$x = \frac{-(-4) \pm \sqrt {(-4)^2- 4(1 \times 1)}}{2(1)}$
$x = \frac{4 \pm \sqrt {16- 4}}{2}$
$x = \frac{4 \pm \sqrt {12}}{2}$
[Note: $\sqrt{12} = \sqrt{3\times4} = \sqrt3\times\sqrt4$ = $2\sqrt3$]
$x = \frac{4 \pm 2\sqrt {3}}{2}$
$x = \frac{4}{2} \pm \frac{2\sqrt3}{2}$
$x = 2 \pm \sqrt3$
