(x-1)(x-2)+(x+3)(x-3)+3x+18=0

Solve the equation: (x-1)(x-2) + (x+3)(x-3) + 3x + 18 = 0

1. Expand:

• (x-1)(x-2) = x^2 - 3x + 2
• (x+3)(x-3) = x^2 - 9

2. Combine: (x^2 - 3x + 2) + (x^2 - 9) = 2x^2 - 3x - 7 Add 3x: 2x^2 - 7 Add 18: 2x^2 + 11

So the equation becomes: 2x^2 + 11 = 0

3. Solve: x^2 = -11/2 Real solutions: none (discriminant negative) Complex solutions: x = ± sqrt(-11/2) = ± i sqrt(11/2) = ± (i √22)/2

Answer:

• No real solutions
• In complex numbers: x = ± i √(11/2) (or ± (i√22)/2)