If 2a+3b+6c=0 then at least one root of the equation ax2+bx+c=0 lies in the interval

f(x)=ax2+bx+cLet F(x)=∫f(x)dx=a3x3+b2x2+cxClearly F(0)=0 and F(1)=a3+b2+c=2a+3b+6c6=0⇒F(0)=F(1)=0There exist at least one point c in between 0 and 1 such that F'(x)=0 or ax2+bx+c=0 for some x∈(0, 1).