Solve in the set $R$ the equation
$$2 \cdot [x] \cdot  \{x\} = x^2 - \frac32 \cdot x - \frac{11}{16}$$
where $[x]$ and $\{x\}$ represent the integer part and the fractional part of the real number $x$, respectively.
Given the equation:
\[ 2 \cdot [x] \cdot \{x\} = x^2 - \frac{3}{2} \cdot x - \frac{11}{16} \]
where \([x]\) and \(\{x\}\) represent the integer part and the fractional part of the real number \(x\), respectively.

Let \(x = a + b\) such that \(a \in \mathbb{Z}\) and \(0 \leq b < 1\). Then \([x] = a\) and \(\{x\} = b\).

Substitute \(x = a + b\) into the equation:
\[ 2ab = (a + b)^2 - \frac{3}{2}(a + b) - \frac{11}{16} \]

Expanding and simplifying the right-hand side:
\[ 2ab = a^2 + 2ab + b^2 - \frac{3}{2}a - \frac{3}{2}b - \frac{11}{16} \]

Subtract \(2ab\) from both sides:
\[ 0 = a^2 + b^2 - \frac{3}{2}a - \frac{3}{2}b - \frac{11}{16} \]

Multiply through by 16 to clear the fraction:
\[ 0 = 16a^2 + 16b^2 - 24a - 24b - 11 \]

Rearrange the equation:
\[ 16a^2 + 16b^2 - 24a - 24b - 11 = 0 \]

Group the terms involving \(a\) and \(b\):
\[ 16a^2 - 24a + 16b^2 - 24b = 11 \]

Complete the square for both \(a\) and \(b\):
\[ 16(a^2 - \frac{3}{2}a) + 16(b^2 - \frac{3}{2}b) = 11 \]

\[ 16\left(a^2 - \frac{3}{2}a + \left(\frac{3}{4}\right)^2 - \left(\frac{3}{4}\right)^2\right) + 16\left(b^2 - \frac{3}{2}b + \left(\frac{3}{4}\right)^2 - \left(\frac{3}{4}\right)^2\right) = 11 \]

\[ 16\left((a - \frac{3}{4})^2 - \left(\frac{3}{4}\right)^2\right) + 16\left((b - \frac{3}{4})^2 - \left(\frac{3}{4}\right)^2\right) = 11 \]

\[ 16(a - \frac{3}{4})^2 - 16\left(\frac{9}{16}\right) + 16(b - \frac{3}{4})^2 - 16\left(\frac{9}{16}\right) = 11 \]

\[ 16(a - \frac{3}{4})^2 + 16(b - \frac{3}{4})^2 - 18 = 11 \]

\[ 16(a - \frac{3}{4})^2 + 16(b - \frac{3}{4})^2 = 29 \]

Divide by 16:
\[ (a - \frac{3}{4})^2 + (b - \frac{3}{4})^2 = \frac{29}{16} \]

Since \(0 \leq b < 1\), we need to check integer values of \(a\) such that the equation holds.

1. **Case \(a = 2\):**
   \[ (2 - \frac{3}{4})^2 + (b - \frac{3}{4})^2 = \frac{29}{16} \]
   \[ (\frac{5}{4})^2 + (b - \frac{3}{4})^2 = \frac{29}{16} \]
   \[ \frac{25}{16} + (b - \frac{3}{4})^2 = \frac{29}{16} \]
   \[ (b - \frac{3}{4})^2 = \frac{4}{16} \]
   \[ (b - \frac{3}{4})^2 = \frac{1}{4} \]
   \[ b - \frac{3}{4} = \pm \frac{1}{2} \]
   \[ b = \frac{3}{4} \pm \frac{1}{2} \]
   \[ b = \frac{5}{4} \text{ or } b = \frac{1}{4} \]

   Since \(0 \leq b < 1\), \(b = \frac{1}{4}\).

   Therefore, \(x = a + b = 2 + \frac{1}{4} = \frac{9}{4}\).

2. **Case \(a = 1\):**
   \[ (1 - \frac{3}{4})^2 + (b - \frac{3}{4})^2 = \frac{29}{16} \]
   \[ (\frac{1}{4})^2 + (b - \frac{3}{4})^2 = \frac{29}{16} \]
   \[ \frac{1}{16} + (b - \frac{3}{4})^2 = \frac{29}{16} \]
   \[ (b - \frac{3}{4})^2 = \frac{28}{16} \]
   \[ (b - \frac{3}{4})^2 = \frac{7}{4} \]
   \[ b - \frac{3}{4} = \pm \sqrt{\frac{7}{4}} \]
   \[ b = \frac{3}{4} \pm \sqrt{\frac{7}{4}} \]

   Since \(\sqrt{\frac{7}{4}} > 1\), there is no valid \(b\) in the range \(0 \leq b < 1\).

3. **Case \(a = 0\):**
   \[ (0 - \frac{3}{4})^2 + (b - \frac{3}{4})^2 = \frac{29}{16} \]
   \[ (\frac{3}{4})^2 + (b - \frac{3}{4})^2 = \frac{29}{16} \]
   \[ \frac{9}{16} + (b - \frac{3}{4})^2 = \frac{29}{16} \]
   \[ (b - \frac{3}{4})^2 = \frac{20}{16} \]
   \[ (b - \frac{3}{4})^2 = \frac{5}{4} \]
   \[ b - \frac{3}{4} = \pm \sqrt{\frac{5}{4}} \]
   \[ b = \frac{3}{4} \pm \sqrt{\frac{5}{4}} \]

   Since \(\sqrt{\frac{5}{4}} > 1\), there is no valid \(b\) in the range \(0 \leq b < 1\).

4. **Case \(a < 0\):**
   If \(a\) is negative, then the left-hand side \(2ab\) is negative, but the right-hand side is positive, leading to a contradiction.

Therefore, the only solution is when \(a = 2\) and \(b = \frac{1}{4}\).

The final answer is \(\boxed{\frac{9}{4}}\).