com.cn/tag/in" target="_blank">int main()
{
double a,b,c,d; double x1,x2; cin>>a>>b>>c; d = b*b-4*a*c; if( d<0 ) { cout<<"erf _ a A ~ - pror"` k { a ! O; } else if ( d==0 ) { x1=x2=-b/(2*a); cout<<x1; } else { x1=-bg J : . G 8 N/(2*a)+sqrt(4*a*c-b*b)/(2*a); x2=-b/(2*a)-sqrt(4*a*c-b*b)/(2*a); cout<<x1<<" "<<x2; } return 0;
}= C r C
回答
sqrt (4*a*c-b*b) 小于0
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