Assume the equation $ax^2+bx+c=0$ has two real roots in the interval $(2,3)$.
Then:
$\Delta=b^2-4ac\ge0$ (the condition for real roots);
$2<-\dfrac{b}{2a}<3; x_m=-\dfrac{b}{2a}$ is the abscissa of the minimum of the parabola $f(x)=ax^2+bx+c$.
$2<-\dfrac{b}{2a}<3\Longrightarrow -6a<b<-4a\Longrightarrow b=-4a-y, y\in(0,2a)$.
$12a+5b+2c>0\Longrightarrow c>-6a-\dfrac{5b}{2}=4a+\dfrac{5y}{2}$.
Results:
$\Delta=b^2-4ac<(-4a-y)^2-4a\cdot\left(4a+\dfrac{5y}{2}\right)=y(y-2a)<0$, contradiction with the condition $\Delta\ge0$.
Hence, the equation $ax^2+bx+c=0$ cannot have two real roots in the interval $(2,3)$.