Number of roots of the equation $ax^2+ bx + c = 0$ in $(1,2)$ 
Let $a, b, c \in R, a \ne 0$ such that $a$ and $4a + 3b + 2c$ have the same sign. Then the
number of roots of the equation $ax^2+ bx + c = 0$ that lie(s) in $(1,2)$ is(are)?

I began by writing that $4a^2+3ab+2ac>0$. I tried finding the sign of $f(1)\cdot f(2)$ to use this condition and Bolzano's theorem. Unfortunately, I couldn't find a way to proceed.

Let's start from $4a^2+3ab+2ac=a(4a+3b+2c)>0$.
Then we have
$8a^2+6ab+4ac>0$
Case 1: $-\frac{b}{2a}<2$
$\implies 8a^2+6ab+b^2 > b^2-4ac$
$\implies \frac{8a^2+6ab+b^2}{4a^2} > \frac{b^2-4ac}{4a^2}$ , since $a^2 > 0, a \in \mathcal{R} \wedge a \neq 0$
$\implies 2+\frac{3b}{2a}+\frac{b^2}{4a^2} > \frac{b^2-4ac}{4a^2}$
$\implies (1+\frac{b}{2a})(2+\frac{b}{2a}) > \frac{b^2-4ac}{4a^2}$
$\implies (2+\frac{b}{2a})(2+\frac{b}{2a}) > (1+\frac{b}{2a})(2+\frac{b}{2a}) > \frac{b^2-4ac}{4a^2}$, since $2+\frac{b}{2a}>0$
$\implies (2+\frac{b}{2a})^2 > \frac{b^2-4ac}{4a^2}$
$\implies (2+\frac{b}{2a})^2 - \left(\frac{\sqrt{b^2-4ac}}{2a}\right)^2 > 0$
$\implies \left(2+\frac{b}{2a}+\frac{\sqrt{b^2-4ac}}{2a}\right)\left(2+\frac{b}{2a}-\frac{\sqrt{b^2-4ac}}{2a}\right) > 0$
$\implies \left(-2-\frac{b}{2a}-\frac{\sqrt{b^2-4ac}}{2a}\right)\left(-2-\frac{b}{2a}+\frac{\sqrt{b^2-4ac}}{2a}\right) > 0$
$\implies \left(-\frac{b}{2a}-\frac{\sqrt{b^2-4ac}}{2a}-2\right)\left(-\frac{b}{2a}+\frac{\sqrt{b^2-4ac}}{2a}-2\right) > 0$
$\implies -\frac{b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a} > 2$ (in which case there is no root in between 1 and 2,since both roots > 2)
OR $\;\;-\frac{b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a} < 2$, in which case both the roots are less than 2. For the second case we need further analysis, can you take it from here?
Also, notice that $f(1)-(b+c)=a$ and $f(2)+(b+c)=4a+3b+2c$ have the same sign.
Case 2: $-\frac{b}{2a}\geq 2$
The roots are $-\frac{b}{2a}\pm\sqrt{(\frac{b}{2a})^2-\frac{c}{a}}$, where the larger root $\geq$ 2, so that we may have at most one (the smaller) root in $(1,2)$, but it needs further analysis.
