Find the roots of the following quadratic equation,if they exist,by the method of completing the square: $2x^{2} + x - 4 = 0$.

(N/A) Given equation: $2x^{2} + x - 4 = 0$.
Step $1$: Divide the entire equation by $2$ to make the coefficient of $x^{2}$ equal to $1$:
$x^{2} + \frac{1}{2}x - 2 = 0 \Rightarrow x^{2} + \frac{1}{2}x = 2$.
Step $2$: Add the square of half the coefficient of $x$ to both sides. The coefficient of $x$ is $\frac{1}{2}$,so half of it is $\frac{1}{4}$. Adding $(\frac{1}{4})^{2}$ to both sides:
$x^{2} + 2(x)(\frac{1}{4}) + (\frac{1}{4})^{2} = 2 + (\frac{1}{4})^{2}$.
Step $3$: Write the left side as a perfect square:
$(x + \frac{1}{4})^{2} = 2 + \frac{1}{16} = \frac{32 + 1}{16} = \frac{33}{16}$.
Step $4$: Take the square root of both sides:
$x + \frac{1}{4} = \pm \sqrt{\frac{33}{16}} = \pm \frac{\sqrt{33}}{4}$.
Step $5$: Solve for $x$:
$x = -\frac{1}{4} \pm \frac{\sqrt{33}}{4} = \frac{-1 \pm \sqrt{33}}{4}$.
Thus,the roots are $x = \frac{-1 + \sqrt{33}}{4}$ and $x = \frac{-1 - \sqrt{33}}{4}$.