Namely: [3(a- 1)]2-4×2×(a2-4a-7)
=9a2- 18a+9-8a2+32a+56
= a2+ 14a+65 = a2+ 14a+49+ 16
=(a+7)2+ 16
(a+7)2≥0
∴(a+7)2+ 16>; 0
No matter what the value of a is, equation 2x? +3(a- 1)x+a? -4a-7=0 must have two unequal real roots.