If $2a+3b+6c=0;a,b,c\in R,$ then find that the equation $a{x}^2+bx+c=0$ has at least one root between 

Given, $2 a+3 b+6 c=0$

$\Rightarrow \frac{a}{3}+\frac{b}{2}+c=0$

Let $f^{\text{'}}\left(x\right)=a{x}^2+bx+c,$

Then, ⁣$f\left(x\right)=\frac{a{x}^3}{3}+\frac{b{x}^2}{2}+cx+d$

Now, $f\left(0\right)=d$ and $f\left(1\right)=\frac{a}{3}+\frac{b}{2}+c+d=0+d$ [from eqn. 1]

Since, $f\left(x\right)$ is a polynomial of three degree, then $f\left(x\right)$ is continuous and differentiable everywhere and $f\left(0\right)=f\left(1\right),$ then by Rolle's theorem $f^{\text{'}}\left(x\right)=0$ i.e., $a{x}^2+bx+c=0$ has atleast one real root between $0$ and $1$.