When are complex conjugates of solutions not also solutions? I've heard that for "normal" equations (e.g. $3x^2-2x=0$), if $(a+bi)$ is a solution then $(a-bi)$ will be a solution as well.
This is because, if we define $i$ in terms of $i^2=-1$ then we might as well define $i^\prime=-i$. Since ${i^\prime}^2=-1$ we find $(a+bi)$ has the same algebraic behaviour as $(a+bi^\prime)$.
So what are non-"normal" equations? When are conjugates not also solutions?
 A: An example of a polynomial with non-real coefficients that has two non-conjugate solutions is 
$$
(x - i) (x - 1) = 0
$$
whose roots are clearly $i$ and $1$. Written out, it looks like
$$
x^2 - (1+i) x + i = 0.
$$
As others have said, I think that the person using the word "normal" meant "equation with real-number coefficients." 
A: If the polynomial has real coefficients, and there is a nonreal root, then its conjugate is also a root. Otherwise, there would be at least one nonreal coefficient.
A: Key Idea $\ $ Conjugation $\rm\:x\mapsto \bar x\:$ preserves $\rm\:\color{#c00}{sums\,\ \&\,\ products}\,$ and  $\rm\:\color{#0a0}{fixes\  coefficients}\in\color{#0a0}{\Bbb R}.\:$ Thus by induction it commutes with polynomials $\rm\  \overline{f(w)} = f(\overline w),\ \ f(x)\in\color{#0a0}{\Bbb R}[x],\ $ having all $\,\rm\color{#0a0}{real}$ coefficients, since such polynomials are compositions of said basic operations. $ $ Explicitly
$$ \begin{eqnarray}
\rm \overline{f(w)}\:
&=&\rm\ \  \overline{a_n w^n +\,\cdots + a_1 w + a_0}\\
&=&\rm\,\ \overline{a_n w^n}\, +\,\cdots + \overline{a_1 w} + \overline a_0\ \ by\ \ \ \color{#c00}{\overline{x+y}\, =\, \overline x + \overline y}\ \ \ \forall\ x,y \in\! \Bbb C\\
&=&\rm\,\  \overline a_n\,  \overline w^n+\,\cdots + \overline a_1\overline w + \overline a_0\ \ by\ \ \ \color{#c00}{\overline{x\, *\, y}\, =\, \overline x\, *\, \overline y}\ \ \forall\ x,y \in\! \Bbb C \\
&=&\rm\,\ a_n\, \overline w^n + \,\cdots + a_1 \overline w + a_0\ \ by\ \ \ \color{#0a0}{\overline a = a}\ \  \forall\ \color{#0a0}a\in \color{#0a0}{\Bbb R}\\
&=&\rm\  f(\overline w)\\
\rm So\ \ \ 0 = f(w)\! \ \Rightarrow\ 0 = \bar 0 = \overline{f(w)}\:& =&\ \rm f(\overline w),\, \ \ \text{i.e. $\ \rm w$ root of $\,\rm f\,\Rightarrow\,\bar w$ root of $\,\rm f$}\quad {\bf QED}
\end{eqnarray}$$
Generally it fails if $f$ has non-real coefficients, e.g. $\,\bar w\,$ is a root of $\,x-w\,$ iff $\,\bar w = w,\,$ i.e. $\,w\in \Bbb R.$
