How to solve a quadratic equation times the unknown index?

Question

$$ e^{-\lambda}*(1+\lambda+\lambda^2/2) = 7/12 $$

I am trying to figure out how to solve it. The answer is ~2.3, can I solve it without trying and error? If yes, how about higher order equation like this?

Answer 1

If you're looking for an analytical solution, I fear there's none, meaning that you must use a numerical method.

For that, I've simply re-written your equation as:

$$\lambda = \ln(\frac{12}{7} \cdot (1+\lambda+\frac{\lambda^2}{2}))$$

You start with a random value (I started with 1), and you create the following series:

$$\lambda_{n+1} = \ln(\frac{12}{7} \cdot (1+\lambda_n+\frac{\lambda_n^2}{2}))$$

These are the values you'll get (using MS-Excel):

$i$	$\lambda_i$
1	1
2	1.455287233
3	1.79581344
4	2.022482542
5	2.161883637
6	2.243518034
7	2.289969499
8	2.315972168
9	2.33039519
10	2.338354703
11	2.342734945
12	2.345141739
13	2.34646307
14	2.347188142
15	2.34758592
16	2.347804112
17	2.347923787
18	2.347989424
19	2.348025423
20	2.348045166

Answer 2

Likely the OP is solving a statistical equation. We convert to a gamma regularized form:

$$e^{-\lambda}(1+\lambda+\lambda^2/2) =Q(3,x)=\frac7{12}\implies \lambda=Q^{-1}\left(3,\frac7{12}\right)$$

and apply inverse gamma regularized as shown here. This is a chi square related distribution quantile of $3$ degrees of freedom. For higher order equations:

$$e^{-\lambda}\sum_{n=0}^{a-1}\frac{\lambda^n}{n!}=x\implies \lambda=Q^{-1}(a,x);0\le x\le1$$