Answer
$x_{1}=\dfrac{-3 + \sqrt{17}}{4}$ and $x_{2} = \dfrac{-3 - \sqrt{17}}{4}$
Work Step by Step
Given $2x^2+3x=1 \longrightarrow 2x^2+3x-1=0$
$a= 2, \ b=3, \ c=-1$
Using the quadratic formula: $\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$ we have:
$\dfrac{-3 \pm \sqrt{3^2-4 \times 2\times (-1)}}{2 \times 2} = \dfrac{-3 \pm \sqrt{9+8}}{4} = \dfrac{-3 \pm \sqrt{17}}{4}$
Therefore the solutions are $x_{1}=\dfrac{-3 + \sqrt{17}}{4}$ and $x_{2} = \dfrac{-3 - \sqrt{17}}{4}$
