Answer
$x_{1}= \dfrac{-5 + \sqrt{17}}{4} $ and $x_{2} = \dfrac{-5 - \sqrt{17}}{4} $
Work Step by Step
Given $2x^2=-5x-1 \longrightarrow 2x^2+5x+1=0$
$a= 2, \ b=5, \ c=1$
Using the quadratic formula: $\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$ we have:
$\dfrac{-5 \pm \sqrt{5^2-4 \times 2\times 1}}{2 \times 2} = \dfrac{-5 \pm \sqrt{25-8}}{4} = \dfrac{-5 \pm \sqrt{17}}{4} $
Therefore the solutions are $x_{1}= \dfrac{-5 + \sqrt{17}}{4} $ and $x_{2} = \dfrac{-5 - \sqrt{17}}{4} $
