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