(N/A) Given equation: $2x^{2} + x + 4 = 0$.
Divide the entire equation by $2$ to make the coefficient of $x^{2}$ equal to $1$:
$x^{2} + \frac{1}{2}x + 2 = 0$.
Rewrite the equation as $x^{2} + \frac{1}{2}x = -2$.
To complete the square,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 $\frac{1}{4}$. The square is $(\frac{1}{4})^{2} = \frac{1}{16}$.
$x^{2} + \frac{1}{2}x + \frac{1}{16} = -2 + \frac{1}{16}$.
$(x + \frac{1}{4})^{2} = \frac{-32 + 1}{16} = -\frac{31}{16}$.
Since the square of a real number cannot be negative,there are no real roots for this equation.