Solve the equation 22x - 3(2^x+2) + 32 = 0.

Recall that $a^{mn} = (a^m)^n$ and $a^{m+n} = a^m \cdot a^n$.

$$2^{2x} - 3(2^{x+2}) + 32 = 0$$ $$(2^x)^2 - 3(2^x \cdot 2^2) + 32 = 0$$ $$(2^x)^2 - 3(4 \cdot 2^x) + 32 = 0$$ $$(2^x)^2 - 12(2^x) + 32 = 0$$

We can factor the quadratic expression.
We need two numbers that multiply to 32 and add up to -12. These numbers are -4 and -8.

$$(y - 4)(y - 8) = 0$$

This gives two possible values for $y$:

$$y - 4 = 0 \implies y = 4$$ $$y - 8 = 0 \implies y = 8$$

Case 1: $y = 4$

$$2^x = 4$$

Since $4 = 2^2$, we have:

$$2^x = 2^2$$ $$x = 2$$

Case 2: $y = 8$

$$2^x = 8$$

Since $8 = 2^3$, we have:

$$2^x = 2^3$$ $$x = 3$$

The solutions for $x$ are 2 and 3.

$$x=2, 3$$