$2 x^{2}+x+4=0$ $\Rightarrow 4 x^{2}+2 x+8=0 \quad$ (Multiplying both sides by 2)
$\Rightarrow 4 x^{2}+2 x=-8$
$\Rightarrow(2 x)^{2}+2 \times 2 x \times \frac{1}{2}+\left(\frac{1}{2}\right)^{2}=-8+\left(\frac{1}{2}\right)^{2} \quad$ [Adding $\left(\frac{1}{2}\right)^{2}$ on both sides $]$
$\Rightarrow\left(2 x+\frac{1}{2}\right)^{2}=-8+\frac{1}{4}=-\frac{31}{4}<0$

But, $\left(2 x+\frac{1}{2}\right)^{2}$ cannot be negative for any real value of $x$.

So, there is no real value of $x$ satisfying the given equation.
Hence, the given equation has no real roots.