If 2C8$\cdot$3C1$=$90C58, what is C?

We may check each possible value of $C$, brute force, quick answer.

Well, it is simpler maybe to check the relation modulo nine, i.e. find possible values for $C$ so that $$ (2+C+8)(3+C+1) = (9+0+C+5+8)\text{ modulo nine.} $$ We get the simpler equation $(C+1)(C+4)=(C+4)$ modulo nine, so $C(C+4)=0$ modulo nine. If one factor is divisible by $3$, the other is not. So we have only two cases, $C=0,9$ or $C=5$, so that the one or the other factor is divisible by $9$. We check the $5$ first,

sage: 258*351
90558

and of course, $298\cdot 391=300\cdot 391-2\cdot 391> 117300-1000$ has too many digits. Later edit: The $0$ is ruled out because (working modulo $100$, we get $1\times 8=8$) the last two digits of $208\cdot 301$ are $08$, not $58$.

Here is an easy way to finish the problem once you got the quadratic:

$$499 C = 2745- 10C^2$$ Now since the RHS is a multiple of $5$ so is the LHS, and hence $5|C$.

Since $C$ is a digit, $C=0$ or $C=5$, but $0$ is not a solution to your equation.

All you have to do is check if $5$ is a solution.

P.S. The solution by dan_fulea becomes much shorter if you look $\pmod{11}$ instead of $\pmod{9}$:

$$(2-C+8)(3-C+1) = (9-0+C-5+8) \pmod{11} \\ (-1-C)(4-C)=(1+C) \pmod{11}\\(1+C)(5-C) = 0 \pmod{11}$$

This means that $$C =10 \pmod{11} \mbox{ or } C=5 \pmod{11}$$ and since $C$ is a digit....

Once you've built that quadratic, you can avoid factoring it by taking it modulo 10, since you know the answer you want is a single digit integer.

$$10C^2+499C−2745\equiv5-C\equiv0\pmod {10}$$

From which C=5 quickly.