Find the value of $a$ such that at least one root of the equation $f(x)=x^2 - (a-3)x + a =0$ is greater than $2$. As the title says.
To solve this problem I took two cases and solved them separately:
1. when the $x$ coordinate of the vertex is greater than $2$ and $f(2)>0$; 
2. when $f(2)<0$.
However I wasn't able to find the answer as the two conditions have no common solution (obviously they won't have because in one case $f(2)>0$ and in the other one $f(2)<0$. 
The inequalities which I got after solving for the two cases separately were: 
1. $a<10$ and $a<7$ 
2. $10 < a$ 

And help would be appreciated. :)

The roots of that equation are
$$
\frac{(a-3) \pm \sqrt{a^2-6a+9-4a}}{2}
$$
so specifically we need
$$
\frac{(a-3) + \sqrt{a^2-6a+9-4a}}{2} > 2 \implies \sqrt{a^2-10a+9} > 7 - a
$$
Here's where I made a mistake at first. If $a < 7$ then we have
$$
a^2 - 10a + 9 > 49 - 14a + a^2 \implies a > 10
$$
so that this route gets us no solutions, now if $a > 7$ we just need to have $\sqrt{a^2 - 10a + 9}$ to be real which mean $a^2 - 10a + 9 > 0$. Notice that the roots to this second polynomial are
$$
\frac{10 \pm \sqrt{100 - 36}}{2} = \frac{10 \pm 8}{2} = 9,1
$$
Now since we have the restriction that $a > 7$ we need to see whether the polynomial is positive or negative in $[7,9]$ and likewise from $[9,\infty)$. Plugging in $a=8$ we see that
$$
8^2-10(8)+9 = -16 + 9 < 0
$$
so $a \not\in [7,9]$. Now checking $a=10$ we have
$$
10^2 -10(10) + 9 = 9 > 0
$$
so that indeed $a > 9$. 
Now we need to do something similar for the other root
$$
\sqrt{a^2-10a+9} < a - 7
$$
Now if $a < 7$ this doesn't make sense since a square root can only be positive. Otherwise if $a > 7$ then we have
$$
a^2 - 10a + 9 < a^2 - 14a + 49 \implies 4a < 40 \implies a < 10
$$
as long as $a^2 - 10a + 9 \ge 0 \implies a \ge 9$
So our other root gives us a solution if $a \in [9,10)$. Now combining these we get that $a \in [9, \infty)$.
Sorry for the issues before
