This problem asks me to find the solutions of a quadratic equation and then take the absolute value of their difference. To find the solutions of a quadratic equation, I can use the quadratic formula: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$, where $a=1$, $b=-5$, and $c=5$ in this case. Plugging in these values, I get $x=\frac{5\pm\sqrt{25-20}}{2}=\frac{5\pm\sqrt{5}}{2}$. So the solutions are $\frac{5+\sqrt{5}}{2}$ and $\frac{5-\sqrt{5}}{2}$. To find the absolute value of their difference, I need to subtract them and take the absolute value of the result. I can write the difference as $\frac{5+\sqrt{5}}{2}-\frac{5-\sqrt{5}}{2}=\frac{2\sqrt{5}}{2}=\sqrt{5}$. The absolute value of $\sqrt{5}$ is just $\sqrt{5}$, since it is a positive number. So the final answer is $\sqrt{5}$. 
Find the absolute value of the difference of the solutions of $x^2-5x+5=0$.