设:(b*b-4ac>0)1,#include "math.h"main(){ float a,b,c,x1,x2; scanf("%f,%f,%f",&a,&b,&c); x1=(-b+sqrt(b*b-4ac))/2a; x2=(-b-sqrt(b*b-4ac))/2a; printf("%f,%f",x1,x2);}2，#include "math.h"main(){ float a,b,c,x1,x2,disc,p,q; scanf("%f,%f,%f",&a,&b,&c); disc=b*b-4*a*c; p=-b/(2*a); q=sqrt(disc)/(2*a); x1=p+q; x2=p-q; printf("%f,%f",x1,x2);}问：为什么第二个对，而第一个是错的?