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

(D) Given equation: $2x^{2} + x + 4 = 0$.
Step $1$: Divide the entire equation by $2$ to make the coefficient of $x^{2}$ equal to $1$:
$x^{2} + \frac{1}{2}x + 2 = 0$.
Step $2$: Move the constant term to the right side:
$x^{2} + \frac{1}{2}x = -2$.
Step $3$: 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:
$x^{2} + 2(x)(\frac{1}{4}) + (\frac{1}{4})^{2} = -2 + \frac{1}{16}$.
Step $4$: Simplify the equation:
$(x + \frac{1}{4})^{2} = -\frac{32}{16} + \frac{1}{16}$.
$(x + \frac{1}{4})^{2} = -\frac{31}{16}$.
Since the square of any real number cannot be negative,$(x + \frac{1}{4})^{2}$ cannot be equal to $-\frac{31}{16}$.
Therefore,there are no real roots for the given quadratic equation.