∴(2x+y)^2-3xy= 1

Let t=2x+y and y=t-2x.

∴t^2-3(t-2x)x= 1

That is 6x 2-3tx+t 2- 1 = 0.

∴△=9t^2-24(t^2- 1)=- 15t^2+24≥0

Solution -2 radical 10/5≤t≤2 radical 10/5.

What is the maximum value of 2x+y? 2 roots 10/5