<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] Using dxiv's hint.
$a^2+b^2 = (a+b)^2 - 2ab\\
8 = (a+b)^2 - 2ab\\
-2ab = 8 - (a+b)^2$
Substitute into
$a+b - 2ab = -4\\
(a+b) + 8 - (a+b)^2  = -4\\
(a+b)^2 -(a+b) - 12 = 0\\
(a+b - 4)(a+b + 3) = 0$
$a+b>0$ so $a+b = 4$
using this value in the equation above
$-2ab = 8 - 16\\
ab = 4$
$(y-a)(y-b) = y - (a+b)y + (ab) = y^2 -4y +4 = (y-2)(y-2)$
$a = 2, b = 2$ and $x = 1$
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