#### Answer

$x_{1}= \dfrac{-3 + \sqrt{19}}{5} $ and $x_{2} = \dfrac{-3 - \sqrt{19}}{5} $

#### Work Step by Step

Given $5x^2+6x-2=0$
$a= 5, \ b=6, \ c=-2$
Using the quadratic formula: $\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$ we have:
$\dfrac{-6 \pm \sqrt{6^2-4 \times 5\times (-2)}}{2 \times 5} = \dfrac{-6 \pm \sqrt{36+40}}{10} = \dfrac{-6 \pm \sqrt{76}}{10} = \dfrac{-6 \pm 2\sqrt{19}}{10} = \dfrac{-3 \pm \sqrt{19}}{5} $
Therefore the solutions are $x_{1}= \dfrac{-3 + \sqrt{19}}{5} $ and $x_{2} = \dfrac{-3 - \sqrt{19}}{5} $