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?

$$z=a+ib\\z^*=a-ib\\\text{while $a,b\in\mathbb{R}$}\\z^2+z^*+\frac{1}{4}=\\(a+ib)^2+(a-ib)+\frac{1}{4}=\\a^2+2iba-b^2+a-ib=-\frac{1}{4}\in\mathbb{R}\\\therefore a^2+2iba-b^2+a-ib\in\mathbb{R}$$
so we can create 2  equations:
$$\begin{cases}
2iba-ib=0
\\[2ex]
a^2-b^2+a=-\frac{1}{4}
\end{cases}$$
we can solve this:$$ib(2a-1)=0\implies b=0\text{ or }a=\frac{1}{2}$$
let's solve for $b=0$ first: $a^2+a=-\frac{1}{4}\implies a=-\frac{1}{2}\\\therefore z=a+ib=-\frac{1}{2}+0i=-\frac{1}{2}$
now let's solve for a=$\frac{1}{2}$: $\frac{1}{2}^2+\frac{1}{2}-b^2=-\frac{1}{4}\implies1=b^2\implies b=\pm1\\\therefore z=a+ib=\frac{1}{2}\pm1i=\frac{1}{2}\pm i$
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