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 works out not so bad.  Move the big radical to the other side and square:
$$\sqrt{x+3}+\sqrt{5-x} = 2\sqrt{15+2x-x^2}-4$$
$$8 +2\sqrt{15+2x-x^2} = 4(15+2x-x^2)-16\sqrt{15+2x-x^2} +16$$
Let $k=\sqrt{15+2x-x^2}$ to get
$$8+2k = 4k^2-16k+16$$
$$2k^2-9k+4=0$$
$$(2k-1)(k-4)=0.$$
That gives you two quadratics to solve $k=4$ yields $x=1$.  $2k=1$ yields two solutions which are extraneous.
