The solution of the inequation `4^(-x+0.5)-7.2^(-x) < 4,x in R` is

Explanation

To solve the inequation \( 4^{-x + 0.5} - 7 \cdot 2^{-x} < 4 \), we can follow these steps:

### Step 1: Rewrite the terms using base 2
We know that \( 4 = 2^2 \), so we can rewrite \( 4^{-x + 0.5} \) as follows:
\[ 4^{-x + 0.5} = (2^2)^{-x + 0.5} = 2^{-2x + 1} = 2^{1 - 2x} \]
Thus, the inequation becomes:
\[ 2^{1 - 2x} - 7 \cdot 2^{-x} < 4 \]

### Step 2: Move 4 to the left side
Rearranging the inequation gives us:
\[ 2^{1 - 2x} - 7 \cdot 2^{-x} - 4 < 0 \]

### Step 3: Substitute \( t = 2^{-x} \)
Let \( t = 2^{-x} \). Then \( 2^{1 - 2x} = 2 \cdot t^2 \) (since \( 2^{-2x} = (2^{-x})^2 = t^2 \)). The inequation now becomes:
\[ 2t^2 - 7t - 4 < 0 \]

### Step 4: Factor the quadratic expression
We need to factor \( 2t^2 - 7t - 4 \). To do this, we can use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 2, b = -7, c = -4 \):
\[ b^2 - 4ac = (-7)^2 - 4 \cdot 2 \cdot (-4) = 49 + 32 = 81 \]
Thus, the roots are:
\[ t = \frac{7 \pm \sqrt{81}}{2 \cdot 2} = \frac{7 \pm 9}{4} \]
Calculating the roots:
\[ t_1 = \frac{16}{4} = 4, \quad t_2 = \frac{-2}{4} = -\frac{1}{2} \]

### Step 5: Set up the inequality
The quadratic \( 2t^2 - 7t - 4 < 0 \) will be negative between its roots. Therefore, we have:
\[ -\frac{1}{2} < t < 4 \]

### Step 6: Substitute back for \( t \)
Since \( t = 2^{-x} \), we have:
\[ -\frac{1}{2} < 2^{-x} < 4 \]
The left part \( 2^{-x} > -\frac{1}{2} \) is always true since \( 2^{-x} \) is always positive. Now we solve the right part:
\[ 2^{-x} < 4 \implies -x < \log_2(4) \implies -x < 2 \implies x > -2 \]

### Step 7: Combine the results
Thus, we have:
\[ x > -2 \quad \text{and} \quad 2^{-x} \text{ can never be } 0 \]
So, the solution set is:
\[ x < 0 \quad \text{and} \quad x > -2 \]
This gives us the interval:
\[ (-2, 0) \]

### Final Answer
The solution of the inequation is:
\[ \boxed{(-2, 0)} \]