What do brackets mean for mod operation? I'm solving equation 5 = (6 * 8 + 9 * b)(mod 10). I tried to use wolframalpha and it gives me answer b = 3. But if I remove brackets around mod 5 = (6 * 8 + 9 * b) mod 10 it makes a plot, and doesn't give me any real answer. I have no idea how to solve this without guessing the b. So I assume there is some meaning behind those brackets?

Okay.
$\pmod n$ means we are doing modulo arithmetic on equivalence classes.
$5 \equiv (6*8 + 9*b)\pmod {10}$ means to find which modulo class $b$ belongs to.
Perhaps a less confusing notation is $5 \equiv_{10} (6*8+9b)$.  The $\pmod {10}$ isn't something you do.  It's a statement about what "universe" of arithmetic you are working in.  And we can solve it:
$5 \equiv_{10} (6*8+9b)$
$5 \equiv_{10} 48 + 9b$
$5 \equiv_{10} 8 + (-1)b$
$-3\equiv_{10} -b$
$3 \equiv_{10} b$
$b \equiv 3 \pmod {10}$.
And without the parenthesis it means the similar but entirely different "gimme the remainder" operation in the "universe" of regualar old arithetic.
$5 = (6*8 + 9b) \mod 10$ means.
The remainder of $(6*8+9b)\div 10$ is $5$
So $48 + 9b = 10n + 5$ for some number $n$
$9b = -43 + 10n$
$9b = 27 + 10(n-6)$
$b = 3 + 10\frac {n-6}9$ for some integer $\frac {n-6}9$.  We don't actually care what $n$ is.  Just that be if $b= 3 + 10k$ we will get that remainder.
So Wolfram is programmed to solve those in different manners.  Even though in practice the results are very very similar.
Note:  $5 = 6*8 + 9b \mod 10$ would be different.
$-43 = (9b \mod 10)$ but $0 \le 9b \mod 10 < 10$ and that can't be $-43$ so this has no solutions.  (Whereas $5 = (6*8 + 9b)\mod 10$ has infinite solutions and $5 \equiv 6*8 + 9b \pmod{10}$ has one solution that is an equivalence class that represents infinitely many equivalent integers.)
