jmc

Since we know that the quotient when we divide $n$ by $d$ is $x$ with a remainder of $7$, we can write $n/d=x+7/d$. Substituting for $n$ and $d$, this gives $$\frac{x^2+2x+17}{2x+5}=x+\frac{7}{2x+5}.$$Multiplying through by $2x+5$ gives

$$\begin{array}{rl}{x}^2+2x+17& =x(2x+5)+7\\ x^2+2x+17& =2x^2+5x+7\\ 0& =x^2+3x-10\\ 0& =(x-2)(x+5).\end{array}$$Thus $x=2$ or $x=-5$. We are given that $x$ must be positive, so we have $x=\boxed{2}$.

To check, we see that $x^2+2x+17=(2{)}^2+2(2)+17=25$, and $2x+5=2(2)+5=9$, and indeed, the quotient when $25$ is divided by $9$ is $x=2$, with a remainder of $7$.