How to solve $(a-x)^{1.4}-bx=0$? I'm trying to solve the following equation algebraically :
$$(1.722-x)^{1.4}-0.565x=0$$
I can find the solution using Matlab using symbolic function solve or an approximation using Taylor series-
the answer is $1.039.$
I have thought of starting from this one $$(1.722-x)^7-(.565x)^5=0$$ but it turns out that there is no algebraic solutions of polynomial equations of degree $\ge5$ by the Abel–Ruffini Theorem.
Is there a way of doing this ? How ? If not, how do I know ?

Another sub-question that came to me while doing my searches is: 
Why is there only one solution provided by Matlab whereas $$(1.722-x)^7-(.565x)^5=0$$ should have $7$ solutions as the equation is of degree $7$?
What about the Abel–Ruffini theorem with an equation of the form $ax^6+bx^3+c$ ? I can solve it using $y=x^3$ and therefore contradicts the Theorem. I suppose I have misinterpreted it.
 A: In integers, your equation is:
$$(1722 - 1000x)^7 - 10^6 (565x)^5 = 0$$
Or:
$$2\cdot(861-500x)^7 - 5^7\cdot 113\cdot x^5 = 0$$
The way to determine if this has a solution in radicals is to calculate the Galois group, which is a subgroup of $S_7$, and determine if it is solvable.  If so, then there is a solution in radicals.
You can do this calculation with the online calculator for Magma:
P<x>:=PolynomialRing(Rationals());
GaloisGroup(2*(861-500*x)^7 - 5^7 * 113 * x^5);
This returns a group of order $5040$.  In other words, the Galois group is exactly $S_7$, and so there is no solution in radicals.  There may be solutions in terms of other functions, but this is a question of much broader scope.

Incidentally, there are indeed seven distinct roots of this equation, but six of them are complex.
A: $a=1.722-x$ -------------TO MAKE LIFE EASIER
$(1.722-x)^{1.4} - .565x=0$ -------here we substitute (1.722-x) with a
$a^{1.4}+.565a-.97293=0$
$a^7+(.565a^5)=(.97293)^5$ ------multiply the powers by 5 to have an integer.
$\ln(a^7)+\ln(.565)+ln(a^5)=ln(.97293^5)$ --- Here, we use natural logarithm- 
also note that ln(.565*a^5) = ln(.565)+ ln(a^5) 
$\ln(a^7)+\ln(.565)+ln(a^5)=ln(.97293^5)$ ---------used the top concept 
$\ln(a^7)+\ln(a^5)=ln(.97293^5)-ln(.565)$----------Put together like terms. 
$\ln(a^7)+\ln(a^5)= 4337/10000 $ ----------ln(.97293^5)-ln(.565)= ~.4337
$70000\ln(a)+50000\ln(a)=4337  --------------\ln(a^7) = 7\ln(a)$
$12000\ln(a)=4337$
$12\ln(a)=4337/10000 -------- \ln(a^12)=\ln(e^{4337/10000})$
$a^12=e^{4337/10000} -------- \ln(e^a)=a$
$a=e^{4337/120000}$
$a=1.0368$
This is the value with no negative or imaginary solution.
If you have any questions, private message me.
A: To begin with, your first two equations cannot be solved algebraically.
Abel–Ruffini Theorem simply states we cannot find radical solutions for the general polynomial of degree 5 or higher.
But we can obviously solve your last problem with simple substitution.  Or consider the following:$$x^n=0,n\ge5$$$$x=0$$
The theorem has nothing to hold you back from finding solutions, it simply holds you back from solving $$ax^5+bx^4+cx^3+dx^2+ex+f$$
The failure of Matlab to find all 7 solutions to your polynomial is not without reason.  Finding such roots is of high difficulty and methods sometimes fail.
I also note that the other 6 roots can be found here, I consider Wolfram|Alpha to be very reliable.
If you can't see all 6 complex roots, just hit the "show more roots" button.
If you are really interested in such solutions, you could try here.  Scroll down for the solution for $x$ and hit that exact form button.
