Solving $\sqrt{x+3}+\sqrt{5-x}-2\sqrt{15+2x-x^2}=-4$ 
Solve the following equation: $\sqrt{x+3}+\sqrt{5-x}-2\sqrt{15+2x-x^2}=-4$

Since real solutions are to be found, the domain of $x$ is $[-3; 5]$.
I immediately found that $15+2x-x^2$ can be factored to $(5-x)(x+3)$, this gave:
$\sqrt{x+3}+\sqrt{5-x}-2\sqrt{(5-x)(x+3)}=-4$
Squaring both sides was too complicated so I tried to substitute $a=\sqrt{5-x}$ for $a \in [0; \sqrt{8}]$ and $b=\sqrt{x+3}$ for $b \in [0; \sqrt{8}]$, this gave the new multivariable equation: $a+b-2ab=-4$
Another thing I noticed was $a^2+b^2=8$, together with the above equation, I got this system of equations: $\left\{ \begin{array}{l}
a + b - 2ab =  - 4\\
{a^2} + {b^2} = 8
\end{array} \right.$
. Solving this by elimination is quite difficult for me.
This problem needs to be solved using algebra so I wonder how do I continue with this or are there any better way to solve this algebraically?


Squaring both sides was too complicated so I tried to substitute 
$a = \sqrt{5−x}, ~b = \sqrt{x + 3}.$ 
This gave the new multivariable equation: $~a+b−2ab=−4$.


$a + b - 2ab = -4.$ 
$a^2 + b^2 = 8.$

This is nice work, so far.
Subtracting the 1st equation above, from the 2nd equation above gives 
$(a^2 + b^2 + 2ab) - (a + b) = 8 - (-4) = 12$.
Let $u = (a + b).$
Then, $u^2 - u = 12 \implies u \in \{4,-3\}.$ 
However, by the constraints, $~~a,b~~$ must each be positive. 
Therefore, $u = -3$ must be rejected. 
Therefore, $a + b = u = 4.$
This implies that $2ab = 8 = a^2 + b^2 \implies (a - b)^2 = 0.$
Therefore, $a = b = 2,~$ since (again), $~~a,b~~$ must each be positive. 
This forces $x = 1$.
