So we have the binomial expression * I might be not using the correct term,english isnt my first language*
$$ \left[1- \frac 34e^{-j2\pi\cdot f} + \frac 18e^{-j4\pi \cdot f} \right]$$
How do I find the roots using this formula $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ If I replace $e^{-j2\pi\cdot f} $with $x$ and $e^{-j4\pi\cdot f}$ with $x^2$ I get that one $x$ is 4 and the other $2$ but those are not the correct solutions...
I know that the correct solution is
$$ \left(1- \frac 12e ^ {-j2\pi\cdot f}\right)\cdot \left(1- \frac 14e^{-j2\pi f}\right)$$
But how to get to there using the formula I said ..
You're trying to factor $$ 1 - \frac34 e^{-2\pi jf} + \frac18 e^{-4\pi jf} $$ You correctly simplify the notation by letting $x=e^{-2\pi jf}$, so the problem becomes factoring $$ 1 - \frac34 x + \frac18 x^2 \tag{$\ast$} $$ You correctly note that you will want to find the roots of this quadratic, because they will correspond to the factors. You correctly found the roots $$ r_1 = 4 \quad\text{and}\quad r_2 = 2 $$ Now the problem is to understand exactly how the roots relate to the factorization. The relation in general is $$ ax^2+bx+c = a(x-r_1)(x-r_2) $$ (I'm guessing here, but I think you were missing the $a$ on the RHS.) In ($\ast$) we see $a=\frac18$, so we should have $$ 1 - \frac34 x + \frac18 x^2 = \frac18(x-4)(x-2) $$ Check that this is correct. To get the exact form of the solution that you know, a little algebra: $$ \frac18(x-4)(x-2) = \frac1{4\cdot 2} (x-4)(x-2) = \frac{x-4}{4}\cdot\frac{x-2}{2} = (\tfrac14 x - 1)(\tfrac12 x - 1) = (1 - \tfrac14 x)(1 - \tfrac12 x)$$ and now substitute $x=e^{-2\pi jf}$.