1. The problem asks to find the square root of 289. 2. The square root of a number $x$ is a value $y$ such that $y^2 = x$. 3. We want to find $y$ such that $y^2 = 289$. 4. Since $17^2 = 289$, the square root of 289 is 17. 5. Therefore, $$\sqrt{289} = 17.$$