If $2a+3b+6c=0,$ then at least one root of the equation $a{x}^2+bx+c=0$ lies in the interval

Let $f^{\prime }(x)=a{x}^2+bx+c$.
Consider the function $f(x)$ obtained by integrating $f^{\prime }(x)$:
$f(x)=\frac{a{x}^3}{3}+\frac{b{x}^2}{2}+cx+d$.
We can simplify this as:
$f(x)=\frac{1}{6}(2a{x}^3+3b{x}^2+6cx+6d)$.
Check the values at $x=0$ and $x=1$:
$f(0)=d$.
$f(1)=\frac{1}{6}(2a+3b+6c+6d)$
Given $2a+3b+6c=0,$ we have:
$f(1)=\frac{1}{6}(0+6d)=d$.
Since $f(x)$ is continuous on $[0,1]$ and differentiable on $(0,1),$ and $f(0)=f(1),$ then according to Rolle's theorem, there exists at least one value $x\in (0,1)$ such that $f^{\prime }(x)=0$.
Thus, $a{x}^2+bx+c=0$ has at least one root in $(0,1)$.