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 __________.

$$x + 1 > 0, x + 1 \neq 1 \implies x > -1, x \neq 0$$

$$\log_{(x+1)}((2x+3)(x+1)) = 4 - \log_{(2x+3)}((x+1)^2)$$

$$\log_{(x+1)}(2x+3) + \log_{(x+1)}(x+1) = 4 - 2\log_{(2x+3)}(x+1)$$

$$\log_{(x+1)}(2x+3) + 1 = 4 - \frac{2}{\log_{(x+1)}(2x+3)}$$

$$t = \log_{(x+1)}(2x+3)$$

$$t + 1 = 4 - \frac{2}{t}$$

$$t^2 - 3t + 2 = 0 \implies t = 1, t = 2$$

$$t = 1 \implies 2x+3 = x+1 \implies x = -2 \quad (\text{rejected})$$

$$t = 2 \implies 2x+3 = (x+1)^2 \implies x^2 + 2x + 1 = 2x + 3$$

$$x^2 = 2 \implies x = \pm\sqrt{2}$$

$$x = \sqrt{2} \quad (x = -\sqrt{2} \text{ rejected})$$

$$S = (\sqrt{2})^2 = 2$$