A function f(x) is given as $$f\left(x\right)\ =\ 2x^2\ +\ 3x\ +\ 1$$ and function g(x) is given as $$g\left(x\right)\ =\ 3x\ +\ 1$$. What is the sum of all the real values of x where $$f\left(g\left(x\right)\right)\ =\ g\left(f\left(x\right)\right)$$?

Given that,

$$f\left(x\right)\ =\ 2x^2\ +\ 3x\ +\ 1$$

$$g\left(x\right)\ =\ 3x\ +\ 1$$

The value of $$f\left(g\left(x\right)\right)$$ is given as,

$$f\left(g\left(x\right)\right)\ =\ 2\left(3x\ +\ 1\right)^2\ +\ 3\left(3x\ +\ 1\right)\ +\ 1\ =\ 2\left(9x^2\ +\ 6x\ +\ 1\right)\ +\ 9x\ +\ 4\ =\ 18x^2\ +\ 21x\ +\ 6$$

$$g\left(f\left(x\right)\right)\ =\ 3\left(2x^2\ +\ 3x\ +\ 1\right)\ +\ 1\ =\ 6x^2\ +\ 9x\ +\ 4$$

Equating f(g(x)) and g(f(x)), we get,

$$18x^2\ +\ 21x\ +\ 6\ =\ 6x^2\ +\ 9x\ +\ 4$$

$$12x^2\ +\ 12x\ +\ 2\ =\ 0$$

We can see that D > 0 for the above equation, and the roots of the equation are real. So, the sum of all the possible values of x is given by,

Sum of the roots  = $$-\dfrac{b}{a}$$ = $$-\dfrac{12}{12}$$ = $$-1$$

The correct answer is option D.