一元二次方程，超简单

Question

x²-8x+1=0
2x²+1=3x
3x²-6x+4=0
x²+6x+4=0

Answer 1

#include <stdio.h>#include <math.h>int main (void){ int a = 1; int b = 2; int c = 3; float delta; float x1; float x2; delta == b*b - 4*a*c; if (delta > 0 ) { x1 = (-b + sqrt(delta))/(2.0*a);//2都改2.0看看 x2 = (-b + sqrt(delta))/(2.0*a); printf("该一元二次方程有两个解, x1 = %f, x2 = %f\n,"x1, x2);} else if (delta< 0 ) { printf("无解\n"); } else (delta=0) { x1 = (-b)/(2.0*a); x2 = x1; printf("该一元二次方程有唯一解，x1 =x2 =%f\n",x1)； //缺参数 }return 0;}

Answer 2

x=4±√15

x=1或x=1/2

x=√21/3-1或x=-√21/3-1

x=-3±√5

本回答被提问者采纳

Answer 3

超简单还需要提问吗？不过是挺简单的。

x = (-b±(b^2-4ac)^(1/2))/2a