Solve the following equations
`x^(2)-3-{x}=0`
where [x] denotes the greatest integer and {x} fractional part.

Answer

To solve the equation \( x^2 - 3 - \{x\} = 0 \), where \(\{x\}\) denotes the fractional part of \(x\), we can follow these steps:

### Step 1: Understand the fractional part
The fractional part \(\{x\}\) is defined as:
\[ \{x\} = x - [x] \]
where \([x]\) is the greatest integer less than or equal to \(x\). Thus, we can rewrite the equation as:
\[ x^2 - 3 - (x - [x]) = 0 \]
This simplifies to:
\[ x^2 - x - 3 + [x] = 0 \]

### Step 2: Rearranging the equation
Rearranging gives us:
\[ x^2 - x + [x] - 3 = 0 \]

### Step 3: Analyze the range of the fractional part
The fractional part \(\{x\}\) is always in the range:
\[ 0 \leq \{x\} < 1 \]
This means:
\[ 0 \leq x^2 - 3 < 1 \]

### Step 4: Solve the inequalities
We can break this into two inequalities:
1. \(x^2 - 3 \geq 0\)
2. \(x^2 - 3 < 1\)

#### Solving the first inequality:
\[ x^2 - 3 \geq 0 \implies x^2 \geq 3 \implies x \leq -\sqrt{3} \text{ or } x \geq \sqrt{3} \]

#### Solving the second inequality:
\[ x^2 - 3 < 1 \implies x^2 < 4 \implies -2 < x < 2 \]

### Step 5: Combine the results
Now we need to find the intersection of the two results:
1. From the first inequality: \( (-\infty, -\sqrt{3}] \cup [\sqrt{3}, \infty) \)
2. From the second inequality: \( (-2, 2) \)

The intersection of these intervals is:
- For \(x \leq -\sqrt{3}\) and \(-2 < x < 2\): The valid interval is \((-2, -\sqrt{3}]\).
- For \(x \geq \sqrt{3}\) and \(-2 < x < 2\): The valid interval is \([\sqrt{3}, 2)\).

### Final Solution
Thus, the solution to the equation is:
\[ x \in (-2, -\sqrt{3}] \cup [\sqrt{3}, 2) \]