$(x+y)^2 = x^2 + y^2 + 2xy = 289 + 2(120) = 289 + 240 = 529$. Therefore, $x+y = \sqrt{529} = 23$.
$(x+y)^2 = x^2 + y^2 + 2xy = 289 + 2(120) = 289 + 240 = 529$। इसलिए, $x+y = \sqrt{529} = 23$।