The sum of all the real values of $$x$$ satisfying the equation $$2^{(x-1)(x^2+5x-50)} = 1$$ is:

We begin with the given exponential equation

$$2^{(x-1)(x^2+5x-50)} = 1.$$

The base $$2$$ is positive and unequal to $$1$$. For a positive base other than $$1$$, an expression of the form $$a^k$$ equals $$1$$ only when its exponent is zero. Therefore we must have

$$(x-1)(x^2+5x-50)=0.$$

This product is zero exactly when at least one of its factors is zero. We analyze each factor in turn.

First factor.

$$x-1=0 \;\;\Longrightarrow\;\; x=1.$$

Second factor.

$$x^2+5x-50=0.$$

To solve the quadratic, compute its discriminant:

$$\Delta = b^2 - 4ac = 5^2 - 4(1)(-50) = 25 + 200 = 225.$$

The square root of the discriminant is

$$\sqrt{\Delta} = \sqrt{225} = 15.$$

Using the quadratic formula,

$$x = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-5 \pm 15}{2(1)}.$$

This yields two real roots:

$$x = \frac{-5 + 15}{2} = \frac{10}{2} = 5,$$

$$x = \frac{-5 - 15}{2} = \frac{-20}{2} = -10.$$

Collecting all real solutions, we have

$$x = -10,\; 1,\; 5.$$

The question asks for the sum of all these real values:

$$\text{Sum} = (-10) + 1 + 5.$$

Perform the additions step by step:

$$-10 + 5 = -5,$$

$$-5 + 1 = -4.$$

Thus the required sum equals $$-4$$.

Hence, the correct answer is Option C.