We consider only the problem of finding the solutions of $(4 x-2)^2-2 (4 x-2) = 15$. You multiplied out correctly. We will look at the equation you got a little later. But first, we do the problem in a more lazy (and efficient) way.
It seems natural to let $u=4x-2$. Then our equation can be rewritten as $u^2-2u-15=0.$ Possibly we can find the solutions by inspection. Or else we can factor the quadratic, obtaining $u^2-2u-15=(u-5)(u+3).$ Note that $(u-5)(u+3)=0$ when $u=5$, when $u=-3$, and for no other value of $u$. It follows that $x$ is a solution of the original equation when $4x-2=-3$ and when $4x-2=5$, and for no other value of $x$.
We now have a couple of linear equations for the roots. To solve $4x-2=-3$, rewrite as $4x=(-3)+2=-1$, and divide both sides by $4$ to get $x=-\dfrac{1}{4}$. Similarly, the equation $4x-2=5$ has the solution $x=\dfrac{7}{4}$.
Instead of factoring $u^2-2u-15$, we could use the Quadratic Formula. For details, please see below.
Another way: Expand, as you did, and bring all terms to the left-hand side. We obtain the equation $16 x^2-24x-7 =0.$ It turns out that the quadratic polynomial on the left can be factored without too much trouble. (Many quadratic polynomials in school exercises can be factored in a simple way. One cannot count on that in the real world!) So we recall the standard formula for solving the quadratic equation $ax^2+bx+c=0$ (where $a \ne 0$). The solutions are $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$ For a derivation of this formula, any source will do. I even have one on Stack Exchange. The formula is so frequently useful that most students remember it. Now calculate, with $a=16$, $b=-24$, and $c=-7$. It turns out that $b^2-4ac=1024=(32)^2$, so the roots are not complicated.
Comment: Let $P(x)$ be a polynomial with integer coefficients. Suppose that the highest degree term is $ax^n$, and the constant term is $c$. Then any rational root of $P(x)=0$ must be of the form $p/q$, where $p$ is a (possibly negative) factor of $c$, and $q$ is a positive factor of $a$. (This result is often called the Rational Roots Theorem.)
So if we are looking for the rational roots of $16x^2-24x-7=0$, the only possible candidates are $\pm\frac{1}{1}$, $\pm\frac{1}{2}$, $\pm\frac{1}{4}$, $\pm\frac{1}{8}$, $\pm\frac{1}{16}$, $\pm\frac{7}{1}$, $\pm\frac{7}{2}$, $\pm\frac{7}{4}$, $\pm\frac{7}{8}$, and $\pm\frac{7}{16}$. We can just try all these, to see which, if any, work. It isn't quite as unpleasant as it looks. However, it is not really a suitable approach to quadratic equations. After all, the roots might not be rational. And the Quadratic Formula is simple to use.