IMPORTANT: If b^x = b^y, then x = y, as long as b ≠ 0, b ≠ 1 and b ≠ -1
For example, if we have 1^x = 1^y, we cannot conclude that x = y, since 1^x equals 1^y FOR ALL values of x and y. For example, 1² = 1³, but we can't conclude that 2 = 3.

So, let's first see what happens when the base (x) equals 0, 1 and -1

If x = 0, then we have: 0^(2(0²) + 4(0) – 6) = 0^(0² + 8(0) + 6)
Simplify: 0^(-6) = 1/(0^6) = 1/0, so 0^(-6) is UNDEFINED, which means x = 0 is not a solution to the equation (of course x = 0 does not change the SUM of the solutions, but it's useful to examine all of the possible considerations)

If x = 1, then we have: 1^(2(1²) + 4(1) – 6) = 1^(1² + 8(1) + 6)
Simplify: 1^0 = 1^15
Evaluate: 1 = 1
So, x = 1 is another solution to the equation

If x = -1, then we have: (-1)^[2(-1)² + 4(-1) – 6] = (-1)^[(-1)² + 8(-1) + 6]
Simplify: (-1)^(-8) = (-1)^(-1)
Evaluate: 1 = -1
So, x = -1 is NOT a solution to the equation

Now let's assume that x ≠ 0, x ≠ 1 and x ≠ -1 and look for other x-values that satisfy the given equation.
Given: x^(2x² + 4x – 6) = x^(x² + 8x + 6)
Since the bases are the same, we can write: 2x² + 4x – 6 = x² + 8x + 6
Rearrange to get: x² - 4x – 12 = 0
Factor to get: (x - 6)(x + 2) = 0
So, x = 6 and x = -2 are also solutions to the equation.

So, the solutions are x = 0, x = 1, x = 6, and x = -2
0 + 1 + 6 + (-2) = 5

Answer: 5