jmc

Let the roots of this polynomial be $r_1$ and $r_2$. Since the sum of the roots of a polynomial $a{x}^2+bx+c=0$ is $-\frac{b}{a}$ and the product of the roots is $\frac{c}{a}$, $r_1+r_2=5$ and $r_1{r}_2=5$. Squaring the first equation results in $r_1^2+2r_1{r}_2+r_2^2=25$.

Notice that $(r_1-r_2{)}^2=r_1^2-2r_1{r}_2+r_2^2$, so the difference of the roots can be obtained by subtracting 4 copies of the product of the roots from the square of their sum: $r_1^2-2r_1{r}_2+r_2^2=r_1^2+2r_1{r}_2+r_2^2-4r_1{r}_2=25-4(5)=5$. Therefore, $|r_1-r_2|=\boxed{\sqrt{5}}$.

We also could have used the quadratic formula to determine that the roots are $\frac{5\pm \sqrt{5}}{2}$, and the positive difference of these roots is indeed $\boxed{\sqrt{5}}$.