(X+3)(X-4)-18=0

To solve the equation \((X+3)(X-4)-18=0\), we will first expand the left side of the equation:

1. Expand the product: \[ (X+3)(X-4) = X^2 - 4X + 3X - 12 = X^2 - X - 12 \]

2. Substitute back into the equation: \[ X^2 - X - 12 - 18 = 0 \] \[ X^2 - X - 30 = 0 \]

3. Now we can solve the quadratic equation \(X^2 - X - 30 = 0\) using the quadratic formula: \[ X = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a = 1\), \(b = -1\), and \(c = -30\).

4. Calculate the discriminant: \[ b^2 - 4ac = (-1)^2 - 4 \cdot 1 \cdot (-30) = 1 + 120 = 121 \]

5. Now use the quadratic formula: \[ X = \frac{-(-1) \pm \sqrt{121}}{2 \cdot 1} = \frac{1 \pm 11}{2} \]

6. This results in two potential solutions: \[ X = \frac{12}{2} = 6 \quad \text{and} \quad X = \frac{-10}{2} = -5 \]

Thus, the solutions to the equation \((X+3)(X-4)-18=0\) are: \[ X = 6 \quad \text{and} \quad X = -5 \]