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(3a-2)x²;+4ax+2ax+2a+2=0 (3a-2)x²;+6ax+2a+2=0 根據題意的 3a-2≠0 Δ=(6a)²;-4(3a-2)(2a+2)>0 36a²;-4(6a²;-4a+6a-4)>0 9a²;-6a²;-2a+4>0 3a²;-2a+4>0恒成立 ∴a≠2/3 即a的取值範圍為a≠2/3
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