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

Let's assume that $$x^{2}+2x-3 = t$$ 

$$9^{x^{2}+2x-3}-4(3^{x^{2}+2x-2})+27=0$$ can be written as $$9^t-4(3^{t+1})+27=0$$

$$3^{2t}-12(3^t)+27=0$$

Let's assume that $$3^t = y$$

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

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

y = 3 or 9 $$\Rightarrow$$ t = 1 or 2

Let's solve when t = 1

$$x^{2}+2x-3 = 1 \Rightarrow x^{2}+2x-4 = 0$$

$$b^2-4ac\ =\ 4+16\ =\ 20$$

Positive, so the equation has real roots.

Product of possible value of x  = -4

Let's solve for t = 2

$$x^{2}+2x-3 = 2 \Rightarrow x^{2}+2x-5 = 0$$

$$b^2-4ac\ =\ 4+20\ =\ 24$$

Positive, so the equation has real roots.

Product of possible value of x = -5

The product of all values = 20