If $2 a+3 b+6 c=0$, then the equation $a x^2+b x+c=0$ has atleast one real root in

Let $f(x)=\frac{1}{3} a x^3+\frac{1}{2} b x^2+c x$ Then, $f(x)$ is a polynomial. So, it is continuous in $R$. Now, $\quad f(0)=0$ and $\quad f(1)=\frac{a}{3}+\frac{b}{2}+c=\frac{2 a+3 b+6 c}{6}$ $\Rightarrow \quad f(1)=0 \quad[\because 2 a+3 b+6 c=0$, given $]$ $\because f(x)$ is a polynomial, so it is differentiable in $R$; so in $(0,1)$. Hence, by Rolle's theorem, there exists atleast one point $x \in(0,1)$, there exists such that $f^{\prime}(x)=0$ $\Rightarrow \quad a x^2+b x+c=0$ Hence, required interval is $(0,1)$.