If $9^{x^2+2x-3} - 4\left(3^{x^2+2x-2}\right) + 27 = 0$, then the product of all possible values of $x$ is

Since $9 = 3^2$, we can write:

$9^{x^2+2x-3} = \left(3^{x^2+2x-3}\right)^2$

Also, $3^{x^2+2x-2} = 3^{(x^2+2x-3)+1} = 3 \cdot 3^{x^2+2x-3}$

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Let $t = 3^{x^2+2x-3}$. The equation becomes:

$t^2 - 4(3t) + 27 = 0$

$t^2 - 12t + 27 = 0$

$(t - 3)(t - 9) = 0$

$t = 3$ or $t = 9$

Both values are positive, which is necessary since $t = 3^{(\text{something})}$ can never be negative or zero.

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When $t = 3$:

$3^{x^2+2x-3} = 3^1$

$x^2 + 2x - 3 = 1$

$x^2 + 2x - 4 = 0 \quad \cdots (i)$

When $t = 9$:

$3^{x^2+2x-3} = 3^2$

$x^2 + 2x - 3 = 2$

$x^2 + 2x - 5 = 0 \quad \cdots (ii)$

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By Vieta's formulas, for any quadratic $x^2 + bx + c = 0$, the product of its roots $= c$.

From equation $(i)$: product of its two roots $= -4$

From equation $(ii)$: product of its two roots $= -5$

The product of all four roots $= (-4) \times (-5) = \boxed{20}$