I have searched through various site's and forums but couldn't find the answer to my problem,
$$z^2-\frac{1}{2}z-\frac{1}{4}=0$$
How will you factorize this As I can't find $2$ numbers that give me $ac = -\frac{1}{4}$ and $b = -\frac{1}{2}$ when added Is there a different approach for such sums.``
The quadratic formula can be used to find the roots of a quadratic equation. $$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$$ for a quadratic of the form $ax^2+bx+c=0$.
So for $a=1,b=-\frac{1}{2},c=-\frac{1}{4}$, we have that
$x=\dfrac{-\left(\frac{1}{2}\right)\pm\sqrt{(-\frac{1}{2})^2-4(1)(-\frac{1}{4})}}{2(1)}=\dfrac{-\frac{1}{2}\pm\sqrt{\frac{1}{4}+1}}{2}$