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?
 A: 
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 $x = 1$.
A: 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.
A: 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 $x = 1$
