If the function f(x)=ax2+bx+c;  a,b,  c ∈R,c≠0,  has the same sign as 'c' then

Given equation is f(x)=ax2+bx+c;  a,b,  c ∈R Now c=f(0)  ≠0  and f(x)  has the same sign as c=f(0)  for all x∈R,  therefore, graph of y=f(x)  does not intersect X-axis at any point.If c>0 then f(x)>0  for all x∈R  and if c<0  then f(x)<0  for all x∈R.This means that f(x)=0  has non-real roots.⇒  b2−4ac<0  ⇒  b2<4ac.