Find the integers $a \ne 0, b$ and $c$ such that $x = 2 +\sqrt3$ would be a solution of the quadratic equation $ax^2 + bx + c = 0$.
1. Given that \( x = 2 + \sqrt{3} \) is a solution of the quadratic equation \( ax^2 + bx + c = 0 \), we need to find integers \( a \neq 0 \), \( b \), and \( c \).

2. Substitute \( x = 2 + \sqrt{3} \) into the quadratic equation:
   \[
   a(2 + \sqrt{3})^2 + b(2 + \sqrt{3}) + c = 0
   \]

3. Calculate \( (2 + \sqrt{3})^2 \):
   \[
   (2 + \sqrt{3})^2 = 2^2 + 2 \cdot 2 \cdot \sqrt{3} + (\sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3}
   \]

4. Substitute \( (2 + \sqrt{3})^2 \) back into the equation:
   \[
   a(7 + 4\sqrt{3}) + b(2 + \sqrt{3}) + c = 0
   \]

5. Expand and group the rational and irrational parts:
   \[
   7a + 4a\sqrt{3} + 2b + b\sqrt{3} + c = 0
   \]
   \[
   (7a + 2b + c) + (4a + b)\sqrt{3} = 0
   \]

6. For the equation to hold, both the rational and irrational parts must separately equal zero:
   \[
   7a + 2b + c = 0 \quad \text{(1)}
   \]
   \[
   4a + b = 0 \quad \text{(2)}
   \]

7. Solve equation (2) for \( b \):
   \[
   b = -4a
   \]

8. Substitute \( b = -4a \) into equation (1):
   \[
   7a + 2(-4a) + c = 0
   \]
   \[
   7a - 8a + c = 0
   \]
   \[
   -a + c = 0
   \]
   \[
   c = a
   \]

9. Therefore, the solutions for \( a \), \( b \), and \( c \) are:
   \[
   a = a, \quad b = -4a, \quad c = a
   \]

10. Since \( a \neq 0 \), we can write the solution as:
   \[
   (a, b, c) = (x, -4x, x) \quad \text{where } x \text{ is a nonzero integer}
   \]

The final answer is \(\boxed{(a, b, c) = (x, -4x, x) \text{ where } x \text{ is a nonzero integer}}\).