If $$\sqrt{50}+\sqrt{128}=\sqrt\text{N}$$​, what is the value of N?

Correct option is A

Given:
​$$\sqrt{50} + \sqrt{128} = \sqrt{N}$$​​
Formula Used:
​$$(\sqrt{a} + \sqrt{b})^2 = a + b + 2\sqrt{ab}$$​​
Solution:
​$$\sqrt{50} + \sqrt{128} = \sqrt{50 + 128 + 2\sqrt{50 \times 128}} \\= \sqrt{178 + 2\sqrt{6400}} \\= \sqrt{178 + 160} \\= \sqrt{338}$$​​
So, N = 338