How do I solve $3n^2+3n-1=0\pmod{5}$? 
How do I solve $3n^2+3n-1=0\pmod{5}$?

I know the answer should be $1  \pmod 5 $ and $3 \pmod5$, but am unsure how to get there. Thanks!

The quadratic formula still applies, as the integers modulo $5$ is a field (where division by $2$ is allowed), like the rational numbers and the real numbers (this is because $5$ is a prime number):
$$
n\equiv\frac{-3\pm\sqrt{3^2+4\cdot3}}{2\cdot3}\\
=\frac{-3\pm\sqrt{21}}{6}
$$
Now, $-3$ is the same as $2$, $21$ is the same as $1$ and $6$ is the same as $1$. The square root (or rather a square root) of $1$ is $1$. So we simplify and get
$$
n\equiv\frac{2\pm1}{1}
$$
which is easily calculated.
