How many $x$ ?

$\sqrt{2x+8} -2\sqrt{x+5} +2 =0$

$\sqrt{2x+8} = 2\sqrt{x+5}-2$

$(\sqrt{2x+8})^2 = (\sqrt{2x+10}-2)^2$

$2x+8 = 4x+24-8\sqrt{x+5}$

$-2x-16 = -8\sqrt{x+5}$

$(-2x-16)^2 = (-8\sqrt{x+5})^2$

$(-2x)^2 - 2(-2x)(16) + (16)^2 = 64x +320$

$4x^2 + 64x +256 = 64x+320$

After solving it, we get $x =4$ and $x = -4$ .

Since $4 > -4,$ the minimum value of $x$ is $\boxed{-4}$