<s>[INST] Solving roots of a polynomial on $\mathbb{Z}_p$ This is probably a simple question. and I would like to work out an example.
How do we solve $x^2 + x + [1] = 0$ over the field $\mathbb{Z}_7$?
I tried a simple case, for example: $[4] x - [3] = 0$, where I find $x = [3][4]^{-1}$ where we find $[4]^{-1} = [2]$ so $x = [6]$
But I get confused in the quadratic case! Any insight would be great!
 [/INST] Suppose $F$ is any field with
$\text{char}(F) \ne 2, \tag 1$
and
$q(x) = x^2 + ax + b \in F[x] \tag 2$
is a quadratic polynomial with coefficients in $F$; we wish to find the zeroes of $q(x)$; short of simply evaluating $q(x)$ for succesive elements $x \in F$ (which in any event will in principle only be guaranteed to succeed when $\vert F \vert < \infty$), what can be done?
Well, it is worth noting that certain well-known, ordinary procedures such as completing the square and its cousin the quadratic formula may be implemented as long as (1) applies.  Indeed, with 
$q(x) = 0 \tag 3$
we may write
$x^2 + ax = -b, \tag 4$
and then add $a^2/4$ to each side; by (1), this is a well-defined element of $F$; we obtain
$\left (x + \dfrac{a}{2} \right )^2 = a^2 + ax + \dfrac{a^2}{4} = \dfrac{a^2}{4} - b; \tag 5$
if then there exists $c \in F$ with 
$c^2 = \dfrac{a^2}{4} - b, \tag 6$
(4) becomes
$\left (x + \dfrac{a}{2} \right )^2 = c^2, \tag 7$
whence
$x + \dfrac{a}{2} = \pm c, \tag 8$
or
$x = -\dfrac{a}{2} \pm c. \tag 9$
So the problem of finding solutions to (3) is tantamount to discovering if $a^2/4 - b$ is a perfect square in $F$; indeed, we may even write (9) in the form 
$x = -\dfrac{a}{2} \pm \sqrt{\dfrac{a^2}{4} - b} = \dfrac{-a \pm \sqrt{a^2 -4b}}{2}, \tag{10}$
familiar to us as the famous "quadratic formula", assuming of course such a $c \in F$ as in (6)  exists.
In the present case of $F = \Bbb Z_7$, the non-zero squares are 
$1 = 1^2 = 6^2, \; 2 = 3^2 = 4^2, \; 4 = 2^2 = 5^2; \tag{11}$
if
$q(x) = x^2 + x + 1, \tag{12}$
we have
$a = b = 1, \tag{13}$
whence
$c^2 = \dfrac{a^2}{4} - b = \dfrac{1}{4} - 1 = 2 - 1 = 1; \tag{14}$
thus
$c = 1, 6, \tag{15}$
and
$x = -\dfrac{1}{2} \pm 1 = 3 \pm 1 = 2, 4; \tag{16}$
both these results are easily checked.
Of course, if we are searching for roots by systematic evaluation of $q(x)$ at different elements of $F$, once we have discovered one of them, such as $2$, we may find the other via dividing $q(x)$ by $x - 2$, or (much less work!) recalling that $-a$ must be their sum, whence the second zero is $-1 - 2 = -3 = 4$ in $\Bbb Z_7$.    
If we encounter a quadratic for which $a^2/4 - b$ is not a perfect square, then such a $q(x)$ will not split over $F$, and we must consider the extension field $F[x]/(q(x))$ to split $q(x)$; but that is another chapter in a long, long story.
</s>