Given that ax2+bx+c=0 has no real roots and a+b+c<0 then

Since the equation ax2+bx+c=0 has no real roots. Therefore, the curve y=ax2+bx+c does not intersect with x-axis. Consequently ϕ(x)=ax2+bx+c c has same sign for all values of x. It is given that a+b+c<0⇒ϕ(1)=a+b+c<0⇒ϕ(x)<0 for all x⇒ϕ(0)<0⇒c<0