当a,b为何值时,方程x²+2(1+a)x+(3a²+4ab+4b²+2)=0有实数根

∵方程有实数根
∴判别式大于等于0
△=[2(1+a)]²-4(3a²+4ab+4b²+2)
=[4(1+2a+a²)]-4(3a²+4ab+4b²+2)
=(4+8a+4a²)-(12a²+16ab+16b²+8)
=4+8a+4a²-12a²-16ab-16b²-8
=-8a²-16ab-16b²+8a-4
=-4(a²+4ab+4b²)-4(a²-2a+1)
=-4(a+2b)²-4(a-1)²
∵(a+2b)²≥0,(a-1)²≥0
∴△≤0
又∵方程有实数根
∴△=0
∴a+2b=0
a-1=0
∴a=1,b=-1/2