#### Answer

$x_{1}= \dfrac{1 + \sqrt{13}}{4}$ and $x_{2} = \dfrac{1 - \sqrt{13}}{4}$

#### Work Step by Step

Given $4x^2-2x-3=0$
$a= 4, \ b=-2, \ c=-3$
Using the quadratic formula: $\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$ we have:
$\dfrac{-(-2) \pm \sqrt{(-2)^2-4 \times 4\times (-3)}}{2 \times 4} = \dfrac{2 \pm \sqrt{4+48}}{8} = \dfrac{2 \pm \sqrt{52}}{8} = \dfrac{2 \pm 2\sqrt{13}}{8} = \dfrac{1 \pm \sqrt{13}}{4}$
Therefore the solutions are $x_{1}= \dfrac{1 + \sqrt{13}}{4}$ and $x_{2} = \dfrac{1 - \sqrt{13}}{4}$