find the base of numeration in which 135/21 has a zero remainder

find the base of numeration in which 135/21 has a zero remainder

Let the base be x

$\frac{x^2+3x+5}{2x+1}=n \qquad n\in Z\\ $

I am going to try trial and error.

x has to be 6 or bigger because of the 5.

If x=6 this becomes

$\frac{36+16+5}{13}=\frac{57}{13}\ne n$

If x=7

$\frac{49+21+5}{14+1}=\frac{75}{15}=5=n$

So base 7 works