Solution:
Substituting $x = y + \frac{1}{2}$ in the equation, we obtain the equation in $y$:
$$8 y^{4} + 4 y^{2} + a - \frac{3}{2} = 0$$
Using the transformation $z = y^{2}$, we get a quadratic equation in $z$:
$$8 z^{2} + 4 z + a - \frac{3}{2} = 0$$
The discriminant of this equation is $32(2 - a)$ which is nonnegative if and only if $a \leq 2$. For $a \leq 2$, we obtain the roots
$$z_{1} = \frac{-1 + \sqrt{2(2 - a)}}{4}, \quad z_{2} = \frac{-1 - \sqrt{2(2 - a)}}{4}$$
For getting real $y$ we need $z \geq 0$. Obviously $z_{2} < 0$ and hence it gives only non-real values of $y$. But $z_{1} \geq 0$ if and only if $a \leq \frac{3}{2}$. In this case we obtain two real values for $y$ and hence two real roots for the original equation. Thus we conclude that there are two real roots and two non-real roots for $a \leq \frac{3}{2}$ and four non-real roots for $a > \frac{3}{2}$. Obviously the sum of all the roots of the equation is $2$. For $a \leq \frac{3}{2}$, two real roots are given by $y_{1} = +\sqrt{z_{1}}$ and $y_{2} = -\sqrt{z_{1}}$. Hence the sum of real roots is $y_{1} + \frac{1}{2} + y_{2} + \frac{1}{2}$ which reduces to $1$. It follows the sum of the non-real roots is also $1$. Thus
$$\text{The sum of nonreal roots} = \begin{cases} 1 & \text{for } a \leq \frac{3}{2} \\ 2 & \text{for } a > \frac{3}{2} \end{cases}$$