Answer
$x_{1} = \dfrac{3+\sqrt{5}}{4}$ and $x_{2}=\dfrac{3-\sqrt{5}}{4}$
Work Step by Step
Given $ \bigg(x-\dfrac{1}{2}\bigg)^2=\dfrac{x}{2} \longrightarrow x^2-2\times \dfrac{1}{2}x+\bigg(\dfrac{1}{2}\bigg)^2 = \dfrac{x}{2} \longrightarrow x^2 -x +\dfrac{1}{4} -\dfrac{x}{2}=0 \\ \longrightarrow x^2-\dfrac{3}{2}x+\dfrac{1}{4}=0$
$a = 1, \ b = -\dfrac{3}{2}, \ c = \dfrac{1}{4}$
Using the quadratic formula $\dfrac{-b \pm \sqrt{b^2-4ac}}{2a} , $ we have:
$\dfrac{-(\frac{3}{2}) \pm \sqrt{(-\frac{3}{2})^2-4\times 1\times (\frac{1}{4}})}{2\times 1} = \dfrac{\frac{3}{2}\pm \sqrt{\frac{9}{4} - 1}}{2} = \dfrac{\frac{3}{2}\pm \sqrt{\frac{5}{4}}}{2} = \dfrac{\frac{3}{2}\pm \frac{\sqrt{5}}{2}}{2} = \dfrac{\frac{3\pm\sqrt{5}}{2}}{2} = \dfrac{3\pm\sqrt{5}}{4}$
Therefore the solutions are $x_{1} = \dfrac{3+\sqrt{5}}{4}$ and $x_{2}=\dfrac{3-\sqrt{5}}{4}$
