Square:

(2x+5) + 4 sqrt(2x+5) sqrt(x+6) +4 (x+6) = 25

so:

4 sqrt(2x+5) sqrt(x+6) = -4-6x.

Now square again:

16 (2x+5)(x+6) = (4+6x)^2

or:

32·x^2 + 272·x + 480 = 36·x^2 + 48·x + 16

Which simplifies to:

4·(x^2 - 56·x - 116)=0

This is a quadratic with roots x=58, and x=-2.

Now we have to substitute each of these back into the original equation

to check that they are roots (as the squarings can introduce spurious

roots).

root(2 58 +5) + 2root(58+6) = 27,

so 58 is not a root of the original equation.

root(2(-2) +5) + 2root((-2)+6) = 1+2 (2)=5

so x=-2 is a root.

Therefore the only solution to the original equation is x=-2.

RonL