Finding an integer-coefficients quadratic equation which has approximate solutions of $e$ and $\pi$. Question : Find a quadratic polynomial $f(x)=a_2x^2+a_1x+a_0\ (a_i\in\mathbb Z)$ which has the following three conditions.
Suppose that $f(x)=0$ gives two real positive solutions $\alpha, \beta.$
Here, note that $2.71828 1\lt e\lt 2.718282$ and $3.141592\lt \pi\lt 3.141593$.
Condition 1 : $2.7182\lt \alpha \lt 2.7183$.
Condition 2 : $3.1415\lt \beta \lt 3.1416$.
Condition 3 : $H:=|a_2|+|a_1|+|a_0|$ is min.
Example : I found the following. 
$1605x^2-9405x+13706=0$ has solutions $x=2.71827\cdots, x=3.14153\cdots$. $H=24716.$
I don't know if this is the answer for the above question.
Motivation : First, I was interested in finding the following polynomials, which are what I've found.
1. $35x^2-83x-33=0$ has a solution $x=2.71828\cdots$.
2. $19x^2-39x-65=0$ has a solution $x=3.14158\cdots$.
3. $45x^2-121x-64=0$ has a solution $x=3.14159\cdots$. $H=230$.
4. $41x^2-47x-257=0$ has a solution $x=3.14159\cdots$. $H=345$.
Then, I got interested in the above question.

I performed an exhaustive search of all quadratic equations $a\,x^2-b\,x+c=0$ with


*

*$1\le a\le 24716$

*$2.7182+3.1415<b/a<2.7183+3.14156$

*$2.7182\times 3.1415<c/a<\min(2.7183\times 3.14156,24716-a-b)$


and found the smallest $H$ corresponds to
$$
1455\,x^2 - 8526\,x + 12425 = 0
$$
with solutions 
$$x_1=2.71829117343,\quad x_2=3.14150264101$$
and
$$
H=22406.
$$
