<s>[INST] 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?
 [/INST] 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.
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