The sum of squares of all the real solutions of the equation $\log_{(x+1)}(2x^2+5x+3) = 4 - \log_{(2x+3)}(x^2+2x+1)$ is equal to ________.

For the logarithms to be defined, we must satisfy the following conditions:
1. Base of the first logarithm: $x+1 > 0 \Rightarrow x > -1$ and $x+1 \neq 1 \Rightarrow x \neq 0$.
2. Base of the second logarithm: $2x+3 > 0 \Rightarrow x > -\dfrac{3}{2}$ and $2x+3 \neq 1 \Rightarrow x \neq -1$.
3. Arguments must be positive: $2x^2+5x+3 > 0$ and $x^2+2x+1 > 0$.
Taking the intersection of all these conditions, the domain of the equation is $x \in (-1, 0) \cup (0, \infty)$.

Now, factorizing the arguments of the logarithms:
$2x^2+5x+3 = (2x+3)(x+1)$
$x^2+2x+1 = (x+1)^2$

Substitute these into the given equation:
$\log_{(x+1)}((2x+3)(x+1)) = 4 - \log_{(2x+3)}((x+1)^2)$

Using the properties of logarithms:
$\log_{(x+1)}(2x+3) + \log_{(x+1)}(x+1) = 4 - 2\log_{(2x+3)}(x+1)$
$\log_{(x+1)}(2x+3) + 1 = 4 - \dfrac{2}{\log_{(x+1)}(2x+3)}$

Let $t = \log_{(x+1)}(2x+3)$. The equation becomes:
$t + 1 = 4 - \dfrac{2}{t}$
$t - 3 + \dfrac{2}{t} = 0$
$t^2 - 3t + 2 = 0$
$(t-1)(t-2) = 0$
$\Rightarrow t = 1$ or $t = 2$

Case 1: $t = 1$
$\log_{(x+1)}(2x+3) = 1$
$2x+3 = x+1$
$x = -2$
This value is rejected because $x = -2$ does not fall in the domain $x > -1$.

Case 2: $t = 2$
$\log_{(x+1)}(2x+3) = 2$
$2x+3 = (x+1)^2$
$2x+3 = x^2+2x+1$
$x^2 = 2$
$x = \pm\sqrt{2}$

Since $x > -1$, $x = -\sqrt{2}$ is rejected. The only valid solution is $x = \sqrt{2}$.

The sum of squares of all the real solutions is $(\sqrt{2})^2 = 2$.

Answer: $2$