The quadratic $x^2+1300x+1300$ can be written in the form $(x+b)^2+c$, where $b$ and $c$ are constants. What is $\frac{c}{b}$?
To write the quadratic in the given form, I need to complete the square. To do that, I need to add and subtract the square of half the coefficient of $x$, which is $\frac{1300}{2}=650$, to the quadratic. So I get $x^2+1300x+1300=(x^2+1300x+650^2)-650^2+1300$. Now I can factor the first term as a perfect square: $(x^2+1300x+650^2)=(x+650)^2$. So the quadratic is $(x+650)^2-422500+1300=(x+650)^2-421200$. Comparing this to $(x+b)^2+c$, I see that $b=650$ and $c=-421200$. Therefore, $\frac{c}{b}=\frac{-421200}{650}=-648$. 