Solve quadratic congruences I have no idea how answer this. How do i solve quadratic congruences.
There is question:
$$
3x^2-2x-4 \equiv0 \mod{(7)}
$$
I don't know how to do. Help please?
 A: You can use the same methods that work for real numbers. Try   factoring, or the quadratic formula. For the latter you'll have to understand how to "divide by $2$" and to find out whether the discriminant has a square root.
Or, since there are just seven possible roots, try them all.
A: One possible option is: since you want the equation to have a solution mod$7$, you can do the case by case type of solution and go through all
$x$ mod$7$ $=0,1,2,3,4,5,6$ 
which as $7$ is not that big, will do the work and you'll see that the equation has no solutions.
What you can also do is say: Let $3x^2 - 2x - 4 = 7k$, for some integer $k$. Then you can solve this equation as if it was in the real numbers, but do note that you need to guarantee that $k,x$ are going to be integers.
A: Hint $\ {\rm mod}\ 7\!:\,\ \left(\,\color{#c00}3(x\!+\!2)^{\large 2} \equiv \color{#c00}2\,\right)^{\!\large 3} \Rightarrow\ {-}(x\!+\!2)^{\large 6} \equiv 1\,$ contra lil Fermat. $ $ Or more conceptually  the discriminant $\,\color{#c00}{\dfrac{2}3}\equiv \dfrac{9}3\equiv 3\,$ isn't square by Euler's Criterion, i.e. $\,a^{\large 2}\equiv 3\,\Rightarrow\, a^{\large 6}\equiv 3^{\large 3}\equiv -1\, \Rightarrow\!\Leftarrow$
