Is $n^2+3n+6$ divisible by 25, where $n$ is a integer? If we want $n^2+3n+6$ to be divisible by $25$, it firstly has to be divisible by $5$.
So, let's take a look at couple of cases:
$n=5k: 25k^2+15k+6$. The remainder is $6$, so it's not divisible by $25$.
$n=5k+1: 25k^2+25k+10$. The remainder is $10$, so it is not divisible by $25$. 
And so on for $n=5k+2, n=5k+3, n=5k+4.$


*

*Is this a good way to do this?


The other way that came to my mind would be following:
Let's say that $n^2+3n+6$ is divisible by $25$: $$n^2+3n+6=25k$$
$$n^2+3n+(6-25k)=0$$ 
If we solve this equation, we get $$n_{1,2}=\frac{-3 \pm \sqrt{5} \sqrt{20k-3}}{2}.$$ But $\sqrt{5}$ is a irrational number, so is $n$ irrational number too. This is contradiction, so it's not divisible by $25$.


*Is this a good way to solve this problem?


Hint: We have
$$
n^2+3n+6=(n+4)^2 \bmod 5
$$
Now look at Bill's answer at this duplicate:
$n^2 + 3n +5$ is not divisible by $121$
