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