(Х+3)(х-4)=12

Let's solve the equation \((x + 3)(x - 4) = 12\).

1. Expand the left side: \[ (x + 3)(x - 4) = x^2 - 4x + 3x - 12 = x^2 - x - 12 \]

2. Set the equation to 0: \[ x^2 - x - 12 = 12 \] \[ x^2 - x - 12 - 12 = 0 \] \[ x^2 - x - 24 = 0 \]

3. Factor the quadratic: We need two numbers that multiply to \(-24\) and add to \(-1\). Those numbers are \(-6\) and \(4\). \[ (x - 6)(x + 4) = 0 \]

4. Set each factor to zero: \[ x - 6 = 0 \quad \Rightarrow \quad x = 6 \] \[ x + 4 = 0 \quad \Rightarrow \quad x = -4 \]

5. Final solutions: \[ x = 6 \quad \text{or} \quad x = -4 \]

Thus, the solutions to the equation \((x + 3)(x - 4) = 12\) are \(x = 6\) and \(x = -4\).