But it IS "normal" Algebra, this section is for Advanced algebra, such as the study of vector spaces and groups.

Since we are solving an equation with "$b$" as the variable, we can re-write it like so:

$(-64j)b^2 + b + (8 - 64j - \sqrt{18}) = 0$

This is a quadratic equation in $b$, with solution:

$b = \dfrac{-1 \pm \sqrt{1 + (4)(64j)(8-64j -\sqrt{18})}}{-128j}$

It is unclear from your problem whether or not $j$ is intended to be a square root of -1.