How to solve the given fourth order algebraic equation for $\beta$?

Question

Actually, I am working to find the exact solution of the non linear ordinary differential equation $f'''+ff''-f'^2=0,$ where the boundary conditions is $f(0)=s, f'(0)=-1+\gamma{f''(0)+\delta{f'''(0)}},f'(\infty)=0$. So, we assume the solution of the form $f(\eta)=A+Be^{-\beta\eta}$. The application of BC above give the value of A and B as $B={\frac{1}{\beta+\gamma\beta^2-\delta\beta^3}}$ and $A=s-{\frac{1}{\beta+\gamma\beta^2-\delta\beta^3}}$. Substituting the assume solution in the above differential we get $A=\beta$. Now using this relation inn the above to get the following fourth order algebraic equation $$\delta\beta^4-(\gamma+\delta{s})\beta^3+(\gamma{s}-1)\beta^2+s\beta-1=0. ...........>(1)$$ Further, the required solution of the given problem in the following closed form which is read as $$f(\eta)=s-\frac{1}{\beta+\gamma\beta^2-\delta\beta^3}+\frac{1}{\beta+\gamma\beta^2-\delta\beta^3}e^{-\beta\eta}.$$ Please help me that how to find the equation (1) for $\beta$.