Generalizing the Quadratic Formula to a Field with Characteristic $\neq 2$ Here is a definition given by my book:

Suppose $f(x)=ax^2+bx+c\in F[x]$ with $2a\in F^\times$, and set $\Delta=b^2-4ac$.
1) If there is a $\delta\in F$ such that $\delta^2=\Delta$, then the roots of $f(x)$ are $\frac{-b\pm\delta}{2a}$.
2) If there is no such $\delta\in F$, then $f(x)$ has no roots in $F$. 

Indeed, this definition seems familiar, and I know that, for polynomials in $\mathbb R[x]$, this can be proven by completing the square. However, how can I generalize it from $\mathbb R[x]$ to $F[x]$ where $F$ can be any field with characteristic $\neq 2$? I'm unsure whether completing the square still works in this case, and the concepts of fields and polynomial arithmetic seem a bit abstract to the mathematically untalented me. Thanks in advance!
 A: Maybe a little elaboration on the given assumptions and conclusions might be helpful. First of all: what is the characteristic of a field (as you have mentioned in the comments that you have not got to charateristics yet).
The characteristic ${\rm char}~F$ of a field $F$ is the least natural number $n$ such that $n\cdot1=0$. If there is no such $n$ we say that $F$ is of characteristic $0$. An elementary exercise shows that $n$ has to be a prime number, which is due to the fact that a field is in particular a so-called integral domain. This is just the term used for structures where $a\cdot b=0$ implies $a=0$ or $b=0$ (an element $x$ for which there exists an element $y$ such that $x\cdot y=0$ is called a zero divisor; an integral domain (and thus a field) has no zero divisors apart from $0$ itself).

Our familiar fields being $\Bbb Q$ and $\Bbb R$ are both of characteristic $0$. On the other hand $\Bbb Z/2\Bbb Z$ is of characteristic $2$ and in particular here we have $0\equiv2$.

Anyway, the notion of the characteristic is not (explictly) needed here. The crucial assumption is in fact given by $2a\in F^\times$. In general $F^\times$ denotes the (multiplicatively) invertible elements of the field $F$. Well, this is nothing else than all non-zero elements by the field axioms. So we have $F^\times=F\setminus\{0\}$ (this notation becomes useful as soon as not all non-zero elements are forced to be invertible, see ring).
We assume that $2a\in F^\times$, so if ${\rm char}~F=2$ we would have $2a\equiv0a=0\in F^\times$ which is a contradiction. Therefore, from our assumption $2a\in F^\times$ it already follows that ${\rm char}~F\neq2$, but that is not important. 
Knowing this, we can actually consider an arbitrary quadratic polynomial $f(x)=ax^2+bx+c\in F[x]$ ($a\neq0$) and continue by completing the square
\begin{align*}
ax^2+bx+c&=0\\
x^2+\frac bax+\frac ca&=0&&;a\neq0\\
x^2+\frac bax+\left(\frac b{2a}\right)^2&=\left(\frac b{2a}\right)^2-\frac ca&&;2a\in F^\times\iff 2a\neq0\\
\left(x+\frac b{2a}\right)^2&=\left(\frac b{2a}\right)^2-\frac ca\\
\left(x+\frac b{2a}\right)^2&=\frac{b^2-4ac}{(2a)^2}\tag1
\end{align*}
So far so good. Now we ran into the next issue: can we take the square root without leaving the field $F$? In general not (consider $F=\Bbb Q$ and take $f(x)=x^2-2$). Answering questions like: what do we need to adjoin in order to "solve" (i.e. write as product of linear factors) this and that polynomial; are there fields, where we can always take the square root of any element (can you think of one?); etc. pp. lead to the interesting field of Galois Theory, but this goes way beyond the scope of this answer.
The assumptions you gave take care of this situation quite easily. Defining the discriminant $\Delta=b^2-4ac$ is the essential step. Now we have two cases: either there is a square root of $\Delta$ or there is not. These are precisely the two cases given. If have an element $\delta\in F$ such that $\delta^2=\Delta$ we can take the square root and the usual quadratic formula emerges from $(1)$; if we have no such element, we cannot go further than $(1)$.

Consider the quadratic polynomial $f(x)=x^2+1\in\Bbb R[x]$. Here we have $a=1$, $b=0$ and $c=1$. Then $\Delta=0^2-1\cdot1=-1$ and, obviously, there is no element $\delta\in\Bbb R$ such that $\delta^2=-1$ (as one should check nevertheless).
  A more exotic example, consider $f(x)=x^3-3\in\Bbb Z/5\Bbb Z$. Here $\Delta=3$ but there is not element $\delta\in\Bbb Z/5\Bbb Z$ such that $\delta^2=3$ as one can check by hand (there are not that many elements in $\Bbb Z/5\Bbb Z$).

