Answer
$x_{1}= 3i$ and $x_{2} = -3i$
Work Step by Step
Given $2x^2+18=0$
$a= 2, \ b=0, \ c=18$
Using the quadratic formula: $\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$ we have:
$\dfrac{-0 \pm \sqrt{0^2-4 \times 2\times 18}}{2 \times 2} = \dfrac{\pm \sqrt{-144}}{4} = \dfrac{\pm \sqrt{144}i}{4} = \dfrac{\pm \sqrt{12^2}i}{4} = \dfrac{\pm 12i}{4} = \pm 3i$
Therefore the solutions are $x_{1}= 3i$ and $x_{2} = -3i$