How to solve quadratic inequalities with two boundaries I was solving a math question on an online test, and the question was asking what is possible values of $x$ if the perimeter of some rectangle tile is between $20$ cm and $54$ cm. I wondered how we would find possible values of $x$ if the area was between $20$ and $54$ cm.
I thought we could use sign analysis for each inequality, but I understood that won't work because we will be having different inequalities in each part and that doesn't make sense.
Inequality looks like this: $20<2x^2+12x<54$
How could we solve quadratic inequalities with two boundaries like this?

The values of $x$ that verify the inequality, will verify the following inequalities at the same time:
$$
\begin{array} 020<2x^2+12x, & 2x^2+12x<54\end{array}
$$
So all we have to do is work with both and we will intersect the solution spaces, getting the solution of the original inequality. Let's start with the first one:
$$
20<2x^2+12x \implies 0<(x+3+\sqrt{19}) ·(x+3-\sqrt{19}) \implies x>-3+\sqrt{19} \  \lor \ x<-3-\sqrt{19}
$$
The second one will have the following result:
$$
2x^2+12x<54 \implies x^2+6x-27<0 \implies -9<x<3
$$
The intersection is all of the $x\in \mathbb{R}\ $ such that:
$$
-9<x<-3-\sqrt{19} \ \lor -3+\sqrt{19}<x<3
$$
