#### Answer

$x_{1}=1$ and $x_{2}= -\dfrac{5}{2}$

#### Work Step by Step

Given $2x^2+3x=5 \longrightarrow 2x^2+3x-5=0$
$a=2, \ b=3, \ c=-5$
Using the quadratic formula: $\dfrac{-\ b\pm \sqrt{b^2-4ac}}{2a}, $ we have:
$\dfrac{- 3\pm \sqrt{3^2-4\times 2\times (-5)}}{2\times 2} = \dfrac{- 3\pm \sqrt{9+40}}{4} = \dfrac{- 3\pm \sqrt{49}}{4} = \dfrac{- 3\pm 7}{4}$
Therefore, the solutions are $x_{1}= \dfrac{- 3 + 7}{4} = \dfrac{4}{4} = 1$ and $x_{2}= \dfrac{- 3 - 7}{4}= \dfrac{-10}{4}=-\dfrac{5}{2}$