The sum of all the real solutions of the equation
$$\log_{(x+3)}{(6x^{2}+28x+30)}=5-2\log_{(6x+10)}{(x^{2}+6x+9)}$$ is equal to

Given: $$\log_{(x+3)}{(6x^{2}+28x+30)}=5-2\log_{(6x+10)}{(x^{2}+6x+9)}$$

$$6x^2 + 28x + 30 = (x +3)(6x + 10)$$ 

$$x^{2}+6x+9 = (x + 3)^2$$

$$\Rightarrow$$ $$\log_{(x+3)}{(x+3)(6x+10)} = 5 - 2\log_{(6x+10)}{(x + 3)^2}$$

$$\Rightarrow$$ $$\log_{(x+3)}{(x+3)} + \log_{(x+3)}{(6x+10)} = 5 - 4\log_{(6x+10)}{(x + 3)}$$

$$\Rightarrow$$ $$ 1 + \log_{(x+3)}{(6x+10)} = 5 - 4\log_{(6x+10)}{(x + 3)}$$

$$\Rightarrow$$ $$ \log_{(x+3)}{(6x+10)} + 4\log_{(6x+10)}{(x + 3)} = 4$$

Let $$ \log_{(x+3)}{(6x+10)} = t$$ $$\Rightarrow$$ $$\log_{(6x+10)}{(x + 3)} = \dfrac{1}{t}$$

$$\Rightarrow$$ $$ t + \dfrac{4}{t} = 4$$

$$\Rightarrow$$ $$t^2 - 4t + 4 = 0$$

$$\Rightarrow$$ $$(t - 2)^2 = 0$$

$$\Rightarrow$$ $$ t = 2$$

$$\therefore$$ $$\log_{(x+3)}{(6x+10)} = 2$$

$$\Rightarrow$$ $$(6x + 10) = (x + 3)^2$$

$$\Rightarrow$$ $$x^2 + 6x + 9 = 6x + 10$$

$$\Rightarrow$$ $$x^2 = 1$$ $$\Rightarrow$$ $$x = \pm 1$$

Both of these values of $$x$$ satisfy the given equation

$$\therefore$$ Sum of real solutions $$= (1) + (-1) = 0$$

Hence, option B is the correct choice