I dont expect the answear, I just want to know if it's possible, and if it is, if it's theres a method for it.
Heres the equation. My goal is to isolate $\lambda$.
$$\frac{\eta}{\lambda} \frac{1}{v} (\epsilon - w) \left[\frac{\alpha + \tau h - \theta \gamma - (1-s_r)(\rho \delta) - s_r i \lambda }{X - \beta} \right] = \frac{X(\alpha + \tau h - \theta \gamma) - \beta[(1-s_r)\rho \delta + s_r i \lambda]}{X - \beta}$$
Being $X = \dfrac{[1 - (1-s_r)\rho h - \varphi (1-h) - \eta (\epsilon - w)]}{v}$
I know it may be hard. But is it possible?
I've tried to isolate, but I ends up having a $\lambda ^2$ on one side, which makes it impossible to isolate, since theres singles $\lambda$ too.
I'm stuck at this part (left side) $\lambda v [ X(\alpha + \tau h - \theta \gamma - \beta [(1-s_r)\rho \delta - s_r i \lambda] + \eta (\epsilon - w) s_r i] $
Notice how theres one inside the [] that I can't get out, therefore I'm not able to isolate.
Any help is much appreciated.
Tag recommendation will also de welcomed. Not sure where to post this exactly.
With OP's clarification that all quantities are real and $\,i\,$ is a variable, not the imaginary unit, let $\mu = s_r i \lambda$, $a = \dfrac{s_r i\eta}{v}$, $b=\alpha + \tau h - \theta \gamma$, $c=(1-s_r)\rho \delta$, then the equation can be written as:
$$ \require{cancel} \begin{align} \color{blue}{\frac{s_r i}{s_r i}}\frac{\eta}{\lambda} \frac{\epsilon - w}{v} \frac{\alpha + \tau h - \theta \gamma - (1-s_r)\rho \delta - s_r i \lambda }{\cancel{X - \beta}} &= \frac{X(\alpha + \tau h - \theta \gamma) - \beta\left((1-s_r)\rho \delta + s_r i \lambda\right)}{\cancel{X - \beta}} \\ \iff\quad\quad \frac{a}{\mu}\left(b-c-\mu\right) &= bX-\beta(c + \mu) \\ ab - a c - a\mu &= bX\mu -\beta c\mu -\beta\mu^2 \\ \underbrace{\beta}_{A} \mu^2 - \underbrace{(bX - \beta c + a)}_{B}\mu + \underbrace{ab - ac}_{C} &= 0 \end{align} $$
The latter is a quadratic equation of the form $A\mu^2 - B\mu+C=0$ which can be solved for $\mu$ with the roots being $\mu_{1,2}=\dfrac{B \pm \sqrt{B^2-4AC}}{2A}$, then reversing the substitution $\lambda_{1,2} = \dfrac{1}{s_r i} \mu_{1,2}\,$.