Complex number quadratic Is the following equation

$$z^2 + z^* + \frac14 = 0$$

where $z$ is a complex number and $z^*$ is its conjugate
completely separate from ordinary quadratic equations? i.e. can I use the discriminant, quadratic formula etc. If not what, what type of equation is this? Can z* be treated independently from z? How is the degree related to the number of roots which is 4 (2 real, 2 complex) I believe.
p.s. which specific topics could I look at to help me understand this further?

Here is a more geometric approach. We have 
$$z^{2}+z^{*} = -\frac{1}{4} \in \mathbb{R}$$
so $z^{2}+z^{*}$ is real. Write this in polar form 
$$r^{2}e^{2i\theta}+re^{-i\theta}=-\frac{1}{4}$$
to obtain the constraint
$$r^{2}\sin(2\theta)-r\sin(\theta)=0$$
so that either $r=0$ (so $z=0$), or $\sin(\theta)=0$ (so $z \in \mathbb{R}$), or 
$$2r\cos(\theta)-1=0 \implies r\cos(\theta)=\frac{1}{2}$$
which we recognise as the condition $\Re(z)=\frac{1}{2}$.
$z=0$ is not a solution; for real $z$ we get the repeated solution $z=-1/2$, and if we say $z=1/2+iy$, the equation becomes:
$$\frac{1}{4}+iy-y^{2}+\frac{1}{2}-iy+\frac{1}{4}=0$$
which simplifies to $y^{2}=1$. We easily verify that $1/2 \pm i$ are both solutions. As you note, this is four roots (with multiplicity) - contrast this with an ordinary quadratic equation, which always has $2$ roots over $\mathbb{C}$.
