Alice and Bob play a game. Alice writes an equation of the form $ax^2 + bx + c =0$, choosing $a$, $b$, $c$ to be real numbers (possibly zero). Bob can choose to add (or subtract) any real number to each of $a$, $b$, $c$, resulting in a new equation. Bob wins if the resulting equation is quadratic and has two distinct real roots; Alice wins otherwise. For which choices of $a$, $b$, $c$ does Alice win, no matter what Bob does?
To determine for which choices of \(a\), \(b\), and \(c\) Alice wins no matter what Bob does, we need to analyze the conditions under which the quadratic equation \(ax^2 + bx + c = 0\) does not have two distinct real roots.

1. **Discriminant Analysis**:
   The quadratic equation \(ax^2 + bx + c = 0\) has two distinct real roots if and only if its discriminant is positive. The discriminant \(\Delta\) of the quadratic equation is given by:
   \[
   \Delta = b^2 - 4ac
   \]
   For the equation to have two distinct real roots, we need:
   \[
   \Delta > 0 \implies b^2 - 4ac > 0
   \]

2. **Bob's Move**:
   Bob can add or subtract any real number \(d\) to each of \(a\), \(b\), and \(c\). This results in a new equation:
   \[
   (a+d)x^2 + (b+d)x + (c+d) = 0
   \]
   The discriminant of this new equation is:
   \[
   \Delta' = (b+d)^2 - 4(a+d)(c+d)
   \]

3. **Discriminant of the New Equation**:
   We need to analyze the conditions under which \(\Delta' \leq 0\) for all possible values of \(d\). Expanding \(\Delta'\):
   \[
   \Delta' = (b+d)^2 - 4(a+d)(c+d)
   \]
   \[
   \Delta' = b^2 + 2bd + d^2 - 4(ac + ad + cd + d^2)
   \]
   \[
   \Delta' = b^2 + 2bd + d^2 - 4ac - 4ad - 4cd - 4d^2
   \]
   \[
   \Delta' = b^2 - 4ac + 2bd - 4ad - 4cd - 3d^2
   \]

4. **Ensuring Non-Positive Discriminant**:
   For Alice to win, \(\Delta' \leq 0\) for all \(d\). This implies that the quadratic expression in \(d\):
   \[
   -3d^2 + (2b - 4a - 4c)d + (b^2 - 4ac) \leq 0
   \]
   must be non-positive for all \(d\). This is only possible if the quadratic expression is identically zero or has no real roots, which implies:
   \[
   2b - 4a - 4c = 0 \quad \text{and} \quad b^2 - 4ac \leq 0
   \]

5. **Solving the Conditions**:
   From \(2b - 4a - 4c = 0\), we get:
   \[
   b = 2a + 2c
   \]
   Substituting \(b = 2a + 2c\) into \(b^2 - 4ac \leq 0\):
   \[
   (2a + 2c)^2 - 4ac \leq 0
   \]
   \[
   4a^2 + 8ac + 4c^2 - 4ac \leq 0
   \]
   \[
   4a^2 + 4ac + 4c^2 \leq 0
   \]
   \[
   a^2 + ac + c^2 \leq 0
   \]
   Since \(a^2 + ac + c^2 \geq 0\) for all real \(a\) and \(c\), the only solution is:
   \[
   a = 0, \quad c = 0
   \]
   Thus, \(b = 0\).

Conclusion:
Alice wins if and only if \(a = b = c = 0\).

The final answer is \( \boxed{ (a, b, c) = (0, 0, 0) } \).