This problem is easy we start : 

Note : \(3x^2+6x+7=3\left(x^2+2x+1\right)+4=3\left(x+1\right)^2+4\Rightarrow\sqrt{3x^2+6x+7}\ge2\)

\(5x^2+10x+14=5\left(x^2+2x+1\right)+9=5\left(x+1\right)^2+9\Rightarrow\sqrt{5x^2+10x+14}\ge3\)

So the left side of the expression \(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}\ge5\)

Come to the right side : \(4-2x-x^2=5-\left(x^2+2x+1\right)=5-\left(x+1\right)^2\le5\)

The thread \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}=5\\4-2x-x^2=5\end{matrix}\right.\Leftrightarrow x=-1\)

So the expression has the only solution x = -1.