Find the all possible values of $a$, such that $4x^2-2ax+a^2-5a+4>0$ holds $\forall x\in (0,2)$ 
Problem: Find the all possible values of $a$, such that
$$4x^2-2ax+a^2-5a+4>0$$
holds $\forall x\in (0,2)$.


My work:
First, I rewrote the given inequality as follows:
$$
\begin{aligned}f(x)&=\left(2x-\frac a2\right)^2+\frac {3a^2}{4}-5a+4>0\end{aligned}
$$
Then, we have
$$
\begin{aligned}
0<x<2\\
\implies -\frac a2<2x-\frac a2<4-\frac a2\end{aligned}
$$
Case $-1:\,\,\,a≥0 \wedge 4-\frac a2≤0$.
This leads,
$$
\begin{aligned}\frac {a^2}{4}>\left(2x-\frac a2\right)^2>\left(4-\frac a2\right)^2\\
\implies \frac {a^2}{4}+\frac {3a^2}{4}-5a+4>f(x)>\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4\end{aligned}
$$
For $f(x)>0$, it is enough to take $\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4>0$ with the restriction $a≥0\wedge 4-\frac a2≤0$.
Case $-2:\,\,\,a≤0 \wedge 4-\frac a2≥0$.
We have:
$$
\begin{aligned}\frac {a^2}{4}<\left(2x-\frac a2\right)^2<\left(4-\frac a2\right)^2\\
\implies \frac {a^2}{4}+\frac {3a^2}{4}-5a+4<f(x)<\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4\end{aligned}
$$
Similarly, for $f(x)>0$, it is enough to take $\frac{a^2}{4}+\frac {3a^2}{4}-5a+4>0$ with the restriction $a≤0\wedge 4-\frac a2≥0$.
Case $-3:\,\,\,a≥0 \wedge 4-\frac a2≥0$.
This case implies, $\left(2x-\frac a2\right)^2≥0$.  This means, $f(x)≥\frac {3a^2}{4}-5a+4$
Thus, for $f(x)>0$, it is enough to take $\frac {3a^2}{4}-5a+4>0$ with the restriction $a≥0\wedge 4-\frac a2≥0$.
Finally, we have to combine all the solution sets we get.

I haven't done the calculation, because I want to make sure that the method I use is correct. Do you see any flaws in the method?

I've tried to answer in a way which skips over some of the tedious calculation so as to make it easier to understand what this question means.
$$f(a)=4x^2-2ax+a^2-5a+4>0$$
Consider, for the moment, $x$ to be a constant. Then we have a quadratic expression in $a$ with roots at (details omitted):
$$a_1=- \sqrt{-3x^2+5x+\tfrac94}+x+\tfrac52$$
$$a_2= \sqrt{-3x^2+5x+\tfrac94}+x+\tfrac52$$
In the range of interest of $x$, it turns out that the expression under the square root is always positive, so we always have two real roots. We can plot the roots as follows:

We want $f(a)>0$ for all $x \in (0,2)$. It is easy to see that this requires that $a$ is less than the minimum value of $a_1$ or $a$ is greater than the maximum value of $a_2$.
We therefore differentiate $a_1$ and $a_2$ with respect to $x$ (details omitted) to find the turning points which are, for $a_1$:
$$x=\tfrac16(5-\sqrt{13}), \;\; a=-\tfrac23(\sqrt{13}-5)$$
and for $a_2$:
$$x=\tfrac16(5+\sqrt{13}), \;\; a=\tfrac23(\sqrt{13}+5)$$
And our full solution is therefore:
$$a<-\tfrac23(\sqrt{13}-5) \text{ and } a>\tfrac23(\sqrt{13}+5) $$
