Solve :
`2^(2x) + 2^(x+2) - 4 xx 2^(3) = 0`

Answer

To solve the equation \(2^{2x} + 2^{x+2} - 4 \cdot 2^3 = 0\), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ 2^{2x} + 2^{x+2} - 4 \cdot 2^3 = 0 \]
First, simplify \(4 \cdot 2^3\):
\[ 4 \cdot 2^3 = 4 \cdot 8 = 32 \]
So, the equation becomes:
\[ 2^{2x} + 2^{x+2} - 32 = 0 \]

### Step 2: Substitute \(2^x\)
Let \(y = 2^x\). Then, we can rewrite \(2^{2x}\) as \(y^2\) and \(2^{x+2}\) as \(4y\) (since \(2^{x+2} = 2^x \cdot 2^2 = 4 \cdot 2^x\)):
\[ y^2 + 4y - 32 = 0 \]

### Step 3: Factor the quadratic equation
Now we need to factor the quadratic equation \(y^2 + 4y - 32 = 0\). We look for two numbers that multiply to \(-32\) and add to \(4\). The numbers \(8\) and \(-4\) work:
\[ (y + 8)(y - 4) = 0 \]

### Step 4: Solve for \(y\)
Setting each factor to zero gives us:
\[ y + 8 = 0 \quad \Rightarrow \quad y = -8 \quad (\text{not valid since } y = 2^x > 0) \]
\[ y - 4 = 0 \quad \Rightarrow \quad y = 4 \]

### Step 5: Substitute back for \(x\)
Now substitute back for \(y\):
\[ 2^x = 4 \]
Since \(4 = 2^2\), we have:
\[ 2^x = 2^2 \]
This implies:
\[ x = 2 \]

### Final Answer
Thus, the solution to the equation is:
\[ \boxed{2} \]