Hi! While solving this program I got a problem, that is- ax^2 + bx + c = 0 and this will be solved by [-b +/- root(b^2 - 4ac)]/2a now this equation will have two solutions, one is exat and another is imaginary. I'm having problem with imaginary portion, if thene is a negetive sign under the root i.r, root(b^2 - 4ac) the program will get error. What to do?

you can do one thing, in the root(b^2 - 4ac) section you will get two thing, one is real and another is imaginary, while calculating the root means Code: d=b*b-4*a*c; x=sqrt(d); after this code add a If-Else i.e, Code: if(d>0) x=sqrt(d); else { d2=-1*d; x=sqrt(d2); } Now while printing the value just add a "i" after the "%d" of imaginary part. i.e, Code: printf("%d i",&something); Hope this will solve your problem!

Here is the code of the complete solution of the problem. Code: #include<stdio.h> #include<conio.h> #include<math.h> void main() { float a,b,c,d,x1,x2,y,z; clrscr(); printf("Enter the value of a: "); scanf("%f",&a); printf("Enter the value of b:"); scanf("%f",&b); printf("Enter the value of c:"); scanf("%f",&c); y=b*b-4*a*c; if(y=>1) { d=sqrt(y); x1=(-b+d)/2*a; x2=(-b-d)/2*a; printf("The solutions are\nX1= %f and\nX2= %f",x1,x2); } else { z=(-1)*y; d=sqrt(z); x1=(-b+d)/2*a; x2=(-b-d)/2*a; printf("The solutions are\nX1= %fi and\nX2= %fi",x1,x2); } getch(); } thanks...