求解ax2+bx+c=0方程的根，其中：a=1,b=2,c=3.
A: a=1;b=2;c=3;d=sqrt(b*b-4*a*c)X1=(-b+d)/(2*a)X2=(-b-d)/(2*a)
B: a=1;b=2;c=3;d=sqrt(b*b-4*a*c)X1=(-b+d)/(2*a)X2=(-b-d)/(2*a)
C: a=1;b=2;c=3;d=sqrt(b*b-4*a*c)X1=(-b+-d)/(2*a)
D: a=1;b=2;c=3;d=sqrt(b*b-4*a*c)X1=(-b+d)\(2*a)X2=(-b-d)\(2*a)
A: a=1;b=2;c=3;d=sqrt(b*b-4*a*c)X1=(-b+d)/(2*a)X2=(-b-d)/(2*a)
B: a=1;b=2;c=3;d=sqrt(b*b-4*a*c)X1=(-b+d)/(2*a)X2=(-b-d)/(2*a)
C: a=1;b=2;c=3;d=sqrt(b*b-4*a*c)X1=(-b+-d)/(2*a)
D: a=1;b=2;c=3;d=sqrt(b*b-4*a*c)X1=(-b+d)\(2*a)X2=(-b-d)\(2*a)
举一反三
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