Except for very few cases, you could not get analytical solutions for $$x^\alpha+ax=b$$ and you will need to use some numerical method such as Newton starting from a "reasonable" guess $x_0$. The method will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ So, for your case $$f(x)=x^\alpha+ax-b$$ $$f'(x)=\alpha x^{\alpha-1}+a$$