Finding $m$ so that the roots of $x^2 - (3m - 1)x + m^2 - 2 = 0$ lie inside the interval $(1, 5)$ 
Find $m$ so that the roots of the equation $x^2 - (3m - 1)x + m^2 - 2 = 0$ lie inside the interval $(1, 5)$.

Having $x_1, x_2 \in (1, 5)$, we get $x_1 + x_2 \in (2, 10)$ and $x_1x_2 \in (1, 25)$. By solving these two inequations I got $m \in \left( \sqrt3, \frac{11}{3} \right)$. However, this result is wrong. By substituting $m$ with $2$, for example, the given conditions are not reached.
So, please, explain what I did wrong and how to get the right solution.
Thank you in advance!

Notice that $\Delta = 5m^2 -6m +9 > 0, \forall m \in \mathbb{R}.$ Hence, 
$x = \dfrac{3m-1 \pm \sqrt{5m^2 -6m +9}}{2} \in \mathbb{R}.$ 
$(1):~~~\dfrac{3m-1 - \sqrt{5m^2 -6m +9}}{2} > 1 \implies (3m-3)^2 > 5m^2 -6m +9 \implies 4m(m -3) > 0 \implies  m < 0 ~~\text{or}~~ m > 3.$ But, only $ m > 3$ satisfies the condition (we squared the inequality!).
$(2):~~~ \dfrac{3m-1 + \sqrt{5m^2 -6m +9}}{2} < 5 \implies 5m^2 -6m +9 <(-3m+11)^2 \implies m^2 -15m +28 > 0 \implies m < \dfrac{15-\sqrt{113}}{2} \approx 2.185 ~~\text{or}~~ m > \dfrac{15+\sqrt{113}}{2} \approx 12.185.$ However, only $m < \dfrac{15-\sqrt{113}}{2}$ is solution of the inequality $(2)$.
Thus, combine the results of $(1)$ and $(2)$, we notice that there is no $m$ such that $x \in (1,5)$.
