Roots of an equation with coefficients in $\mathbb{Z}_{19}$ 
In the field $\mathbb{Z_{19}}$, consider the equation $x^2+\bar{5}x-\bar{7}$. Let $r_1$ and $r_2$ be the roots of the equation. If the polynomial $P(x) = x^2+\bar{5}-\bar{7}$ is reducible, then the roots belong to $\mathbb{Z_{19}}$. If it is irreducible, then they belong in the Galois field $F = GF(19,P(x))$. Compute $r_1^2 + r_2^2$.

This is what I'm working on. In the class of $\mathbb{Z_{19}}$, we have the elements ${\bar 0,\bar 1,\bar 2,\bar 3,\bar 4...\bar {16},\bar {17},\bar {18}}$. So basically, I believe I have to find two solutions to the polynomial $P(x) = x^2+\bar{5}x-\bar{7}$. Or, in other words, $ x^2+\bar{5}x-\bar{7} \equiv 0$ (mod $19)$. I could use the trial and error method, but for future problems, what if I work with a different field with more elements in its class, such as $\mathbb{Z_{29}}$ for example? There must be an efficient way of finding these roots without doing the trail and error approach.
 A: (Edited to incorporate some comments made after question was clarified.)
Notice that if $r_1 $ and $r_2$ are the two roots of the quadratic in a splitting field, then
$$ (x - r_1)(x - r_2) = x^2 + \bar 5 x - \bar 7.$$
Therefore,
$$ r_1 + r_2 = - \bar 5, \ \ \ \ \ r_1 r_2 = - \bar 7.$$
From here, we have
$$ r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2r_1 r_2 = (- \bar 5)^2 - 2 ( - \bar 7) = \bar 1 \in \mathbb Z_{19}.$$

In fact, the quadratic does not split in $\mathbb Z_{19}$. One easy way to see this is to complete the square, i.e. write it as $(x + \overline{12})^2 + \bar 1$. Now note that $- \bar 1$ is not a quadratic residue mod $19$, since $-1$ is the ninth power of the generator of the group of units mod $19$, and nine is odd. So the splitting field for the quadratic is the finite field of order $19^2$, and the roots $r_1$ and $r_2$ live in this splitting field.
A: Bhaskara's formula works, in the sense that a solution exists if and only if $\sqrt{b^2-4ac}$ exists in $\mathbb{Z}_{19}$. In other words, if and only if $b^2-4ac$ is a quadratic residue modulo $19$. Do you think you can take it from here?
