Question 733581
in an eqn ax^2+bx+c=0 the discriminant=b^2-4ac 
& the solns are [-b+-sqrt(b^2-4ac)]/2a
now if 1)b^2=4ac then there is one solution i.e -b/2a
    if 2)b^2>4ac then there are two solutions i.e [-b+-sqrt(b^2-4ac)]/2a
but if 3)b^2<4ac then there are no real solutions asthe sqrt will have no real value
now here in x^2-5x+6=0          a=1,b=-5& c=6
so b^2-4ac =1
so there are two solutions by 2) above 3 & 2 ans