已知x的一元二次方程x²+2(m+1)+3m²+4mn+4n²+2=0有实数根,则3m²+2n²的值为多少

4(m+1)^2-4(3m^2+4mn+4n^2+2)≥0
m^2+2m+1-3m^2-4mn-4n^2-2)≥0
-2m^2-4mn-4n^2+2m-1≥0
2m^2+4mn+4n^2-2m+1≤0
m^2+4mn+4n^2+m^-2m+1≤0
(m+2n)^2+(m-1)^2≤0
平方和非负,所以
(m+2n)^2=0,（m-1)^2=0
m=-2n,m=1
m=1,n=11/2
3m²+2n²=3*1^2+2*(-1/2)^2=7/2
3m²+2n²的值为 7/2