How can I solve for mod to get a value? I want to solve the following part involving mod:
1 = -5(19) (mod 96)

Apparently, this mod in brackets (mod 96) here is different from the mod that I know e.g. its not the remainder value that you get by dividing.
What of kind of mod is it and how can I solve it step by step to get the result which is 77?
Update:
Okay, so I am trying to find what's 5 inverse mod 96.
By following euclidean algorithm approach here's what I am doing:
Step1: Find GCD of 5 and 96
96 = 5(19) + 1

which becomes 1 = 96-5(19) when expresses in 1 term
Step 2: Take mod (96) both sides
So I will have left:
1 = -5(19) (mod 96)

That's from where I need to solve it to get 5 inverse (mod 96).   
 A: The notation $$ a\equiv b \pmod n $$
means that $a$ and $b$ are in the same residue class modulo $n$. If you're more used to $\bmod$ as a binary operator, this is the same as saying
$$ a\bmod n = b\bmod n$$
though we have to remember that the $\bmod$ here is one that always produces a result in $[0,n)$ even for a negative argument, such that, as will be relevant here, $(-19)\bmod 96 = 77$.

When you're being asked to find a multiplicative inverse of $5$ modulo $96$, what this means is to find an $x$ such that
$$\tag1 5x\equiv 1 \pmod{96} $$
Through the previous steps of the solution you have reached the knowledge that
$$\tag2 1 \equiv -5 \cdot 19 \pmod{96} $$
which is almost the same as what you're looking for, except that


*

*The $1$ is on the other side -- but that doesn't matter, because the definition of $\equiv$ is clearly symmetric in $a$ and $b$.

*There is a $-5$ instead of a $5$ in $\text{(2)}$. But we can get a $5$ instead by rembering (basic aritmetic) that $(-5)\cdot 19 = 5\cdot(-19)$.


So, since we know $\text{(2)}$ is true, we also know
$$\tag3 5\cdot(-19) \equiv 1 \pmod{96} $$
which tells us that $-19$ is a solution for $x$ in $\text{(1)}$.
All that is left to do is find the canonical representive of the residue class that contains $-19$, by adding or subtracting some multiple of $96$ to get it into the range $[0,96)$:
$$ -19 \bmod 96 = 77 $$
In other words, since changing one factor by a multiple of $n$ doesn't change the residue modulo $n$, we also know that
$$ 5\cdot 77 \equiv 1 \pmod{96}$$
which says that $5$ and $77$ are multiplicative inverses modulo $96$.
A: Based on the update of your post, you are looking for $x$ for which 
$$5x\equiv 1\pmod{96}.$$
This $x$ is called the multiplicative inverse of $5$ in mod $96$. Now, we have
$$1\equiv (-19)5\pmod{96}$$ which means that $$5(-19)\equiv 1\pmod{96}.$$ This means any $x$ such that $$x\equiv -19\pmod{96}$$ is the inverse of $5$ in mod $96$.  Observe that $$-19\equiv 77\pmod{96}.$$  Hence, all solutions of $x$ are given by$$x\equiv 77\pmod{96}.$$
A: $1 \equiv -5(19) \mod 96$ is equivalent to $1 \equiv -95 \mod 96$ by multiplying out the brackets. 
For your question it's equivalent to going round a clock with $96$ hours on it - each time you reach $96$th hour it goes back to $0$ and starts again. 
This can also be extended to negative numbers, and adding multiples of $96$. Hence:
$$1 \equiv -5(19) \mod 96 \equiv 1 \mod 96$$
