令t=√(x^2+2x+2p)>=0
则：x^2+2x=t^2-2p
代入原方程得：t^2+2t-p^2-2p=0
得：(t+1)^2=(p+1)^2
即t=p or t=-p-2
1）若方程无实根,
则：√(x^2+2x+2p)=p,无实根,
p=0,且x^2+2x+2p-p^2=0无实根,但此时(x+1)^2=(p-1)^2,得：x=p-2 or x=-p ,不符
√(x^2+2x+2p)=-p-2,无实根,
-p-2-2
或 -p-2>=0,且x^2+2x+2p-(p+2)^2=0无实根,delta=4[1-2p+(p+2)^2]=4(p^2+2p+5)