2x^2+2xy+5y^2+2x+4y+1=0
2x^2+(2y+2)x+(5y^2+4y+1)=0
这个方程有解则判别式大于等于0
(2y+2)^2-8(5y^2+4y+1)>=0
4y^2+8y+4-40y^2-32y-8>=0
36y^2+24y+4
解一道一元二次方程题目已知实数x,y满足关系式2x^2+2xy+5y^2+2x+4y+1=0,求x,y的值
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