$\sqrt{40-9x}-2\sqrt{7-x}=\sqrt{-x}$ In MAO 1991,

Find $2x+5$ if $x$ satisfies $\sqrt{40-9x}-2\sqrt{7-x}=\sqrt{-x}$

My attempt,
I squared the the equation then I got $144x^2+1648x+4480=144x^2-1632x+4624$, which results $x=-9$, and $2x+5=-13$.
I want to ask is there another way to solve this question as my method is very tedious. Thanks in advance.

Substitution $3.5-x=t$ seems to help because $-x$ and $7-x$ are symmetric around the point $3.5$ you get
$$\sqrt{9t+8.5}=\sqrt{t-3.5}+2\sqrt{t+3.5}\\9t+8.5=t-3.5+4t+14+4\sqrt{t^2-3.5^2}\\4t-2=4\sqrt{t^2-3.5^2}\\(2t-1)^2=4(t^2-3.5^2)\\4t^2-4t+1=4t^2-4\cdot 3.5^2\\4t=4\cdot 3.5^2+1\\4t=(2\cdot 3.5)^2+1\\4t=50\\t=12.5\\x=-9$$
