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To expand the expression, start by dealing with each part separately. First, distribute \(3\) to \((2x-3)(x+2)\): \[ 3(2x-3)(x+2) = 3[(2x)(x) + (2x)(2) - (3)(x) - (3)(2)] = 3(2x^2 + 4x - 3x - 6) = 3(2x^2 + x - 6) = 6x^2 + 3x - 18 \] Next, expand \(-2(x-3)(x+3)\): \[ -2((x)(x) + (x)(3) - (3)(x) - (3)(3)) = -2(x^2 - 9) = -2x^2 + 18 \] Now combine both expanded parts: \[ (6x^2 + 3x - 18) + (-2x^2 + 18) = 6x^2 - 2x^2 + 3x - 18 + 18 = 4x^2 + 3x \] So, the final result is: \[ 4x^2 + 3x \]