Solving $12x \equiv 20 \pmod{38}$ $12x \equiv 20 \pmod{38}$
$gcd(12,38)=2$ using Euclidean Algorithm.
There is a solution since $2|20$.
Use the Extended Euclidean Algorithm $2=12*-3 +1*38$
Then $20=2*10=12*-3*10+1*38*10$ so $x=-3*10=-30$ is a solution.
General solution: $x=-30+19t$ where t is an integer.
My question is: is the above correct or should $-30$ be changed to a positive number i.e. $8$

Assume $x(t) = -30 + 19t$ is a general solution, so that $12x(t) \equiv 20 \ (\text{mod }38)$ for all integers $t$.
Then $g(t) = x(t) + n\cdot 38$ is also a general solution for all $n \in \mathbb{Z}$ since
$$12g = 12x + 12\cdot n\cdot 38 \equiv 12x \ (\text{mod }38) \text{, and } 12x\equiv 20 \ (\text{mod } 38)$$
For all integers $t$. 
Your solutions are equivalent!
