Quadratic equation with parameter and conditions on the roots For which values of $m$ the equation
$$(2m+1)x^2-(4m+2)x+m-1=0$$
has two real roots $x_1$ and $x_2$ that satisfy the condition: $x_1<x_2<2$?
I found that $$m\in\left(-\frac{1}{2}; -\frac{1}{3}\right).$$ Is it correct?

The discriminant is $$\Delta=b^2-4ac=(4m+2)^2-4(2m+1)(m+1)=8m^2+20m+8$$ To get real roots, $m$ must satisfy $$\Delta>0$$
which reduces as $$2m^2+5m+2>0$$
and has for solution $m>-1/2$ or $m<-2$. The roots are then $$x_{2,1}=\frac{-b\pm\sqrt{\Delta}}{2a}=\frac{4m+2\pm\sqrt{8m^2+20m+8}}{4m+2}$$
with $x_1=1-\frac{\sqrt{8m^2+20m+8}}{4m+2}$ and $x_2=1+\frac{\sqrt{8m^2+20m+8}}{4m+2}$.
If we want $x_1<x_2$, then
$$1-\frac{\sqrt{8m^2+20m+8}}{4m+2}<1+\frac{\sqrt{8m^2+20m+8}}{4m+2}$$
which yields $m>-1/2$, consistent with what we found above.
Now, we also want $x_2<2$, so we solve
$$\frac{\sqrt{8m^2+20m+8}}{4m+2}<1$$
and we find $m>1$ or $m\le-2$.
Combining all these conditions, if we want real roots $x_1$ and $x_2$ that verify $x_1<x_2<2$, we need
$$m\in\left]-\infty;-2\right[\cup\left]1;+\infty\right[$$
Of course, you should always check these results, and also interpret them graphically. For instance, using Mathematica, the command

Manipulate[Plot[(2*m + 1)*x^2 - (4*m + 2)*x + m - 1, {x, -1, 3}], {m, -10, 10}]

creates a nice parametric plot and you can observe the behaviour of the solutions as they approach the boundaries.
