已知m、n是一元二次方程x²-3x+1=0的两根,求代数式2m²+4n²-6n+1990

m、n是一元二次方程x²-3x+1=0的两根
由韦达定理得：
m+n=3
mn=1
m²-3m+1=0
m²=3m-1
n²=3n-1

2m²+4n²-6n+1990
=2(3m-1)+4(3n-1)-6n+1990
=6m-2+12n-4-6n+1990
=6m+6n-6+1990
=6(m+n)-6+1990
=18-6+1990
=12+1990
=2002