(viii) abs(x–1)·abs(x+2) = 3
  Given x ∈ ℝ:
    If abs(x–1)·abs(x+2) = 3:
      If x-1 ≥ 0 ∧ x + 2 ≥ 0:
        x ≥ 1
        (x–1)·(x+2) = 3
        x^2+x-5 = 0
        1^2−4·1·(–5) ≥ 0 ∧ ( x = (–1+sqrt(1^2−4·1·(–5)))/2 ∨ x = (–1–sqrt(1^2−4·1·(–5)))/2 ) [lemma]
        x = (–1–sqrt(21))/2 ∨ x = (–1+sqrt(21))/2
      x ≥ 1 ⇒ x = (–1–sqrt(21))/2 ∨ x = (–1+sqrt(21))/2
      If x-1 ≤ 0 ∧ x + 2 ≤ 0:
        x ≤ 1 ∧ x ≥ –2
        (–x+1)·(–x–2) = 3
        x^2+2·x–x–2–3 = 0
        x^2+x–5 = 0
        1^2−4·1·(–5) ≥ 0 ∧ ( x = (–1+sqrt(1^2−4·1·(–5)))/2 ∨ x = (–1–sqrt(1^2−4·1·(–5)))/2 ) [lemma]