$z=\frac{3-4i±\sqrt{(3-4i)^2-4\cdot 1\cdot (2-4i)}}{2}$
Now, $(3-4i)^2-4\cdot 1\cdot (2-4i)=-15-16i=1^2+(4i)^2-2\cdot 1\cdot 4i=(1-4i)^2$
So, $z=\frac{3-4i±(1-4i)}{2}=1$ or $2-4i$
$4w^2=-15-8i=(1)^2+(4i)^2-2\cdot 1\cdot 4i=(1-4i)^2$
$2w=±(1-4i)$
Alternatively, let $-\frac{15}{4}-2i=(a-ib)^2$
So, $-\frac{15}{4}-2i=a^2-b^2-i2ab$
$a^2-b^2=-\frac{15}{4}$ and $2ab=2$
$(a^2+b^2)^2=(a^2-b^2)^2+4a^2b^2=(a^2-b^2)^2+(2ab)^2=\frac{289}{16}$
$a^2+b^2=\frac{17}{4}$
$a^2=\frac{1}{4},b^2=4$
So,$a=±\frac{1}{2},b=±2$ as $a,b$ are of same sign.
The square root of $(3-4i)^2-4\cdot 1\cdot (2-4i)=-15-16i$ can be calculated in the same way.