Answer
{$\dfrac{-3 \pm \sqrt{41}}{4}$}
Work Step by Step
Given: $2x^2+3x-4=0$
or, $x^2+\dfrac{3}{2}x=2$
Add $(\dfrac{3}{4})^2$ on both sides to make a perfect square.
Thus,
$x^2+\dfrac{3}{2}x+(\dfrac{3}{4})^2=2+(\dfrac{3}{4})^2$
$x^2+2(x)(\dfrac{3}{4})+\dfrac{9}{16}=\dfrac{41}{16}$
or, $(x+\dfrac{3}{4})^2=\dfrac{41}{16}$
Apply Square root property.
$x=-\dfrac{3}{4}\pm \dfrac{\sqrt{41}}{4}$
Hence,we have $x=${$\dfrac{-3 \pm \sqrt{41}}{4}$}