If the two positive numbers whose difference is $12$ and whose AM exceeds the GM by $2$, then the numbers are

Let numbers be $a>b$ $ a-b=12 $ $ \frac{a+b}{2}-\sqrt{ab}=2 $ $ \Rightarrow a+b-2\sqrt{ab}=4 $ $ \Rightarrow (a+b)^2=(4+2\sqrt{ab})^2 $ $ \Rightarrow a^2+2ab+b^2=16+4ab+16\sqrt{ab} $ Using $a-b=12$ $ (a-b)^2+4ab=16+4ab+16\sqrt{ab} $ $ 144+4ab=16+4ab+16\sqrt{ab} $ $ \Rightarrow 128=16\sqrt{ab} $ $ \Rightarrow \sqrt{ab}=8 $ $ \Rightarrow ab=64 $ Now, $ a+b=4+2\cdot8=20 $ Solving, $ a=16,\ b=4 $