How to solve this quadratic congruent equation by inspection I found a systematic way (c.f. How to solve this quadratic congruence equation) to solve all congruent equations of the form of $ax^2+bx+c=0\pmod{p}$, or to determine that they have no solution.
But I wonder if there is some easy way to find solutions of simple quadratic congruent equations by inspection, for example, for this equation $$x^2+x+47=0\pmod{7}.$$
My textbook gave the solutions $x\equiv 5\pmod{7}$ and $x\equiv 1\pmod{7}$ directly, without any proof. So I think there must be some easier way to inspect the solutions.
Any help will be appreciated.
 A: The good old quadratic equation works just fine if $p\neq2$. If $a\not\equiv 0\pmod{p}$ and $b^2-4ac$ is a quadratic residue mod $p$, then the solutions to the quadratic congruence
$$ax^2+bx+c\equiv0\pmod{p},$$
are precisely
$$-\frac{b\pm\sqrt{b^2-4ac}}{2a}.$$
In this particular case we have $p=7$ and $a\equiv b\equiv1\pmod{7}$ and $c\equiv47\equiv5\pmod{7}$. Then 
$$b^2-4ac\equiv2\equiv3^2\pmod{7},$$
so the congruence has the two solutions
$$-\frac{1+3}{2}=5\qquad\text{ and }\qquad-\frac{1-3}{2}=1.$$

On the other hand, if you want to solve it purely by inspection, note that there are only $7$ possible solutions to check. Clearly $x=0$ is not a solution, and plugging in $x=1$ yields the first solution. The product of the solutions is congruent to $47\equiv5\pmod{7}$, so the other solution is $x=5$.
A: Servaes gave a good general method for solving quadratic equations modulo a prime.  
I will elaborate on my comment about how in this particular case the answer could be found by inspection.  
Adding or subtracting a multiple of $p$ to or from any of the coefficients $a,b,c$
does not materially change the equation $ax^2+bx+c\equiv0\mod p.$ 
In fact,   $x^2+x+47$ begs to be reduced modulo $7$ to $x^2+x+5.$ 
$x^2+x+5$ doesn't have rational roots, but you could add or subtract any multiple of $p$ to the $x$-coefficient $1$,
yielding polynomials such as $x^2+8x+5, x^2-6x+5,$ etc., 
and $x^2-6x+5$ is easily recognized as $(x-1)(x-5)$.  
