If the equation ax2+bx+c=0,a,b,c∈R have non-real roots, then

Since the roots of ax2 + bx + c = 0 are nonreal, so, f(x) = ax2 + bx + c will have same sign for every value of x. Hence,    f(0)=c,f(1)=a+b+c,f(−1)=a−b+c    f(−2)=4a−2b+c⇒     c(a+b+c)>0,c(a−b+c)>0,c(4a−2b+c)>0