$\sqrt{40-9x}-2\sqrt{7-x}=\sqrt{-x}$ 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.
 A: We can construct some nice solutions after having known the answer a posteriori. 
Put $x=-y$ where $y>0$ (as the quantity under square root must be positive). 
Hence
$$\sqrt{40+9y}-2\sqrt{7+y}=\sqrt{y}$$
from which it can be seen that $y=9$ (i.e. $x=-9$) satisfies the equation. 
Alternatively, put $u=x+9$ (i.e. $x=u-9$) in the original equation, giving
$$\begin{align}
\sqrt{40-9(u-9)}-2\sqrt{7-(u-9)}&=\sqrt{9-u}\\
\sqrt{121-u}-2\sqrt{16-u}&=\sqrt{9-u}\end{align}$$
Since $121, 16, 9$ are perfect squares, we try putting $u=0$ to make the square roots disappear and in doing this we find that the equation is satisfied. Hence the solution is $u=0$ i.e. $x=-9$.
A: Substitution $3.5-x=t$ seems to help because $-x$ and $7-x$ are symmetric around the point $3.5$ you get
$$\sqrt{9t+8.5}=\sqrt{t-3.5}+2\sqrt{t+3.5}\\9t+8.5=t-3.5+4t+14+4\sqrt{t^2-3.5^2}\\4t-2=4\sqrt{t^2-3.5^2}\\(2t-1)^2=4(t^2-3.5^2)\\4t^2-4t+1=4t^2-4\cdot 3.5^2\\4t=4\cdot 3.5^2+1\\4t=(2\cdot 3.5)^2+1\\4t=50\\t=12.5\\x=-9$$
A: An $x$ that solves your equation also solves the system
\begin{align} 
40&&-9x&-a^2&=0\\
7&&-x&-b^2&=0\\
&&-x&-c^2&=0\\
&&a-2b&-c&=0\\
\end{align}
If we subtract the second and the third equation we get 
$$-b^2+c^2+7=0.$$
It turns out that 
$$-9(-b^2+c^2+7)=-(3b-4c)(-3b-4c)-(-7c^2+63).$$
Therefor one solution is $c^2-9=0$ and $3b-4c=0$. Both solutions to $c$ work. If we pick $c=3$ we get $x=-9$ from the third equation and $b=4$ which makes $x=-9$ consistent with the second equation.
