Thank you for explaining. I think the solution is only x=6 . If x=-8/3, the left side is √(-25/3) + √(-4/3) = 7i/(√3) = (7/3)×(√3)i , but the right side is √(-3) = (√3)i Thus, even if the domain is complex number, x=-8/3 does NOT satisfy the given equation. (If my calculation is wrong or there are any miss-typings, please inform me. I will check it.)
Getting rid of three square roots would entail squaring three times, yielding an octic. That can't be the intention. So let's guess that it comes out easy because everything is an integer. So each radicand is a square integer. The two LHS radicands differ by 7, so they are 9 and 16, yielding 2x=12 and x=6. And this turns out to work.
