General formula for $(x+a)^{\frac 1x} =b$

Note that : this problem can be also solved by Lambert W-function, but I prefer this solution because it is similar to "Completing the square method" and the formula looks as if it was an elementary solution, also note that Lambert W-functions must have logarithm in its formula to solve this problem.

Solution:
This solution depends on the method that I call it "Completing the super square" because its steps looks like the known method of solving the quadratic equation.

By starting with.
$x + a = b^x$
Let  $y = x + a$
Substitute $y$ in the equation
$y = b^{y-a} \Rightarrow y = \frac {b^y}{b^a}  \Rightarrow b^a y = b^y \Rightarrow \frac {1}{b^a y} = b^{-y}$
Now powering both sides to $\frac {1}{b^a y}$ for eliminating $y$ in the right side and making same power same base for the left side.
$(\frac {1}{b^a y})^{\frac {1}{b^a y}} = (b^{-y} )^{\frac {1}{b^a y}}$

$(\frac {1}{b^a y})^{\frac {1}{b^a y}} = b^{-y*\frac {1}{b^a y}} = b^{\frac {-1}{b^a} }$
Getting the super square root of both sides
$\sqrt {(\frac {1}{b^a y})^{\frac {1}{b^a y}}}_{s} = {\sqrt {b^{\frac {-1}{b^a} }}}_{s}$
Where $\sqrt {(...)}_{s}$  is super square root, and because the base of the left side equal its power then it is easy to get super square root of it as below.
$\frac {1}{b^a y} = {\sqrt {b^{\frac {-1}{b^a} }}}_{s}$
$y=\frac {1}{b^a {\sqrt {b^{\frac {-1}{b^a} }}}_{s}}$
$x + a =\frac {1}{b^a {\sqrt {b^{\frac {-1}{b^a} }}}_{s}}$

Finally the general formula is
$$x = -a + \frac {1}{b^a {\sqrt {b^{(\frac {-1}{b^a}) }}}_{s}}$$
Example to test the formula:
$(x + 2)^{1/x} = 2$

Solution:
$x = -2 + \frac {1}{2^2 {\sqrt {2^{(\frac {-1}{2^2}) }}}_{s}} = -2 + \frac {1}{4 {\sqrt {2^{(\frac {-1}{4}) }}}_{s}} = -2 + \frac {1}{4 {\sqrt {\frac {1}{2^{(\frac {1}{4})}}}}_{s}}$
By calculating super square root values we obtain two real values.
${\sqrt {\frac {1}{2^{(\frac {1}{4})}}}}_{s} =\frac {1}{16} = 0.0625$ …… (because $0.0625^{0.0625}=0.8409=\frac {1}{2^{(\frac {1}{4})}}$ )
And
${\sqrt {\frac {1}{2^{(\frac {1}{4})}}}}_{s} = 0.8067$  …… (because $0.8067^{0.8067} = 0.8409 = \frac {1}{2^{(\frac {1}{4})}}$) 
$x_{1}=-2 + \frac {1}{4*\frac {1}{16}} = -2 + 4 = 2$

$x_{2}=-2 + \frac {1}{4*0.8067}=-2 + 0.3099 = -1.6901$