So i need to solve for x in the equation floor(x/2) * ceil(2/x) = 17.
If we assume x is an integer, then x/2 * 2/x = 17 implies that x has no integral value.
If x is not an integer, then let x = k + m where k is an integer but m belongs to (0, 1).
Now I have, floor(k+m/2) * ceil(2/k+m) = 17.
Does it make any sense? Can we solve this equation in this way?
As mentioned in comments already, you seem to have confused $\left\lfloor\dfrac{x}{2}\right\rfloor$ with $\dfrac{\lfloor x\rfloor}{2}$. These are not the same.
We have the product of two integers equaling $17$. As $17$ is prime this can only happen in one of four ways: $17\cdot 1,~1\cdot 17,~(-17)\cdot (-1),$ and $(-1)\cdot (-17)$
So, the question becomes "When can $\left\lfloor\dfrac{x}{2}\right\rfloor = 17$ and does $\left\lceil \dfrac{2}{x}\right\rceil = 1$ in any of those instances?"
One should quickly see (by looking at a graph or otherwise) that $\left\lfloor\dfrac{x}{2}\right\rfloor = 17$ happens whenever $x\in[34,36)$ and every $x$ in that range has $\left\lceil\dfrac{2}{x}\right\rceil = 1$ so these are all solutions.
Move to the next case: "When can $\left\lfloor\dfrac{x}{2}\right\rfloor = 1$ and does $\left\lceil \dfrac{2}{x}\right\rceil = 17$ in any of those instances?"
Well, $\left\lfloor\dfrac{x}{2}\right\rfloor = 1$ when $x\in[2,4)$ but in this range we still have $\left\lceil \dfrac{2}{x}\right\rceil = 1$ so no solutions exist of this second form.
Complete the problem by investigating the cases of $(-17)\cdot (-1)$ and $(-1)\cdot (-17)$.