m^2+5=(m+1)^2
m=2
delta=b^2(a-c)^2-(a^2+b^2)(b^2+c^2)=0
so b^4+2acb^2+a^2c^2=0
b^2=-ac
x=-2
so 12m+2n=1
4m-4n=5
so m=1/4
n=-1
SOS~~`紧急...数学题...(有关1元2次方程的~~~)谢谢!~
m^2+5=(m+1)^2
m=2
delta=b^2(a-c)^2-(a^2+b^2)(b^2+c^2)=0
so b^4+2acb^2+a^2c^2=0
b^2=-ac
x=-2
so 12m+2n=1
4m-4n=5
so m=1/4
n=-1