Can this algebraic equation of degree 5 be solved? I have the following algebraic equation of degree 5 which I would like to solve for $x \in \mathbb{R}$:
$$f(x) =ax^3 +bx^2 + cx + d \text{ with } a \in \mathbb{R}_{>0},\; b,c,d,w,z \in \mathbb{R}$$
$$0 \stackrel{!}{=} f'(x) \cdot (f(x)-w) + x - z$$
so when you put it together gives:
$$0 \stackrel{!}{=} 3 a^2 x^5 + 5abx^4 + 2(2ac + b^2 )x^3 + 3(ad+bc-aw) x^2 + (2 b d+c^2+1-2 b w)x+(c d-c w-z)$$
I know that for arbitrary algebraic equations of degree 5 or higher there is no solution formula (Proof: Jacobson, Nathan (2009), Basic algebra 1 (2nd ed.), Dover, ISBN 978-0-486-47189-1, p. 211) But my equation seems to have a lot more structure. 
So is there a solution formula for this type of equation?
Background of this question
I'm currently trying to find out if you can calculate the points on a cubic function with shortest distance to a given point. By looking at graphs I'm sure this equation can have two solutions and it can't have 4 or more (although I don't know how to prove it).
(I'm pretty sure you can find those solutions quite well with Newtons method, but I want to know if you can directly calculate them.)
 A: Unfortunately, your quintics are still too general: Consider $f(x)=x^3$ and $[w,z]=[0,1]$. The resulting equation is $3x^5+x-1=0$ and PARI claims the Galois group of this polynomial is the full $S_5$:
 ? polgalois(3*x^5+x-1)
 %1 = [120, -1, 1, "S5"]

In fact, this is not very surprising -- the polynomial resulting from distance-minimization can be almost arbitrary. If you look at the coefficients in the fully expanded form, you'll see that $a$, $b$ and $c$ can be used to produce arbitrary coefficients at $x^5$, $x^4$ and $x^3$. The next one, $x^2$, can be controlled fully by the difference $(d-w)$, while the constant term can be adjusted arbitrarily by choice of $z$. The only one which is not arbitrary is $x$; its coefficient is determined by the preceding ones.
A: Quintic equations are solvable through radicals if they meet the lengthy criteria described here. If not, they can be solved using Bring radicals. Either way, there's no general solution, as proven by Abel and Galois two centuries ago.
