x2-x-42=(x+a) (x-7)

To solve the equation \( x^2 - x - 42 = (x + a)(x - 7) \), we first expand the right-hand side.

1. Expand the right-hand side: \[ (x + a)(x - 7) = x^2 - 7x + ax - 7a = x^2 + (a - 7)x - 7a \]

2. Now we can set both sides of the equation equal: \[ x^2 - x - 42 = x^2 + (a - 7)x - 7a \]

3. We can eliminate \(x^2\) from both sides: \[ -x - 42 = (a - 7)x - 7a \]

4. Rearranging the equation gives: \[ -x - (a - 7)x = -7a + 42 \] \[ (-1 - (a - 7))x = -7a + 42 \] \[ (-a + 6)x = -7a + 42 \]

5. For the equation to hold for all \(x\), the coefficients must be equal:

   • The coefficient of \(x\) on the left is \( -a + 6 \), and the constant on the right is \( -7a + 42 \).

   This gives the following two equations: \[ -a + 6 = 0 \quad \text{(1)} \] \[ -7a + 42 = 0 \quad \text{(2)} \]

6. Solve equation (1): \[ -a + 6 = 0 \implies a = 6 \]

7. Now substitute \(a = 6\) into equation (2) to verify: \[ -7(6) + 42 = -42 + 42 = 0 \]

Both equations confirm that \(a = 6\).

Therefore, the value of \(a\) that satisfies the original equation is \( \boxed{6} \).