Answer
$x=\frac{1}{2}$
Work Step by Step
$Find$ $all$ $real$ $solutions$ $of$ $the$ $quadratic$ $equation:$
$4x^2-4x+1=0$
Use the Quadratic Equation Formula: $x = \frac{-b \pm \sqrt {b^2-4ac}}{2a}$
$4x^2-4x+1$
$a = 4, b = -4, c = 1$
$x = \frac{-(-4) \pm \sqrt {(-4)^2- 4(4 \times 1)}}{2(4)}$
$x = \frac{4 \pm \sqrt {16- 16}}{8}$
$x = \frac{4 \pm \sqrt {0}}{8}$
$x= \frac{4}{8}$
$x=\frac{1}{2}$
