The sum of all possible values of $x$ satisfying the equations $2^{4 x^{2}}-2^{2 x^{2}+x+16}+2^{2 x+30}=0$, is

We need to find all possible values of $x$ that satisfy:

$2^{4x^2} - 2^{2x^2 + x + 16} + 2^{2x + 30} = 0$

This looks complicated, but there's a beautiful pattern hidden here.

---

Let's break down each term using exponent rules:

For the first term: $2^{4x^2} = 2^{2 \cdot 2x^2} = (2^{2x^2})^2$

For the middle term: $2^{2x^2 + x + 16} = 2^{2x^2} \cdot 2^{x + 16}$

For the last term: $2^{2x + 30} = 2^{2x} \cdot 2^{30}$

Key insight: We can rewrite $2^{2x + 30}$ as $(2^{x + 15})^2$ because $2x + 30 = 2(x + 15)$

---

Let's substitute our rewritten terms. Notice that $x + 16 = (x + 15) + 1$, so:

$2^{2x^2 + x + 16} = 2^{2x^2} \cdot 2^{x + 15} \cdot 2^1 = 2 \cdot 2^{2x^2} \cdot 2^{x + 15}$

Our equation becomes:

$(2^{2x^2})^2 - 2 \cdot 2^{2x^2} \cdot 2^{x + 15} + (2^{x + 15})^2 = 0$

This is a perfect square! Remember the formula: $a^2 - 2ab + b^2 = (a - b)^2$

Here, $a = 2^{2x^2}$ and $b = 2^{x + 15}$

---

Our equation becomes:

$(2^{2x^2} - 2^{x + 15})^2 = 0$

Important property: If any number squared equals zero, then that number itself must be zero.

Therefore: $2^{2x^2} - 2^{x + 15} = 0$

This gives us: $2^{2x^2} = 2^{x + 15}$

---

When we have $2^a = 2^b$, then $a = b$ (since exponential functions are one-to-one).

So: $2x^2 = x + 15$

$2x^2 - x - 15 = 0$

---

We need to factor $2x^2 - x - 15 = 0$

We need two numbers that multiply to $(2)(-15) = -30$ and add to $-1$. Those numbers are $-6$ and $+5$ (since $-6 \times 5 = -30$ and $-6 + 5 = -1$)

$2x^2 - x - 15 = 2x^2 - 6x + 5x - 15$

$2x(x - 3) + 5(x - 3) = (2x + 5)(x - 3) = 0$

Solutions: $2x + 5 = 0$ or $x - 3 = 0$

Therefore: $x = -\tfrac{5}{2}$ or $x = 3$

---

The sum of all possible values is:

$-\tfrac{5}{2} + 3 = -\tfrac{5}{2} + \tfrac{6}{2} = \tfrac{1}{2}$

Answer: $\tfrac{1}{2}$