How would one solve the following equation? This equation is giving me a hard time.
$$e^x(x^2+2x+1)=2$$
Can you show me how to solve this problem algebraically or exactly? I managed to solve it using my calculator with one of its graph functions. But I would like to know how one would solve this without using the calculator.
Highly appreciated,
Bowser.

The answer given by Desmos for intersection of the two curves $y=e^x$ and $y=\frac {2}{(x+1)^2}$ is  $\color{red}{x=0.249}$. Now we have
$$(x+1)^2=2e^{-1}\iff x^2+2x+1=2(1-x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{24}-\frac{x^5}{60}+O(x^6))$$ hence $$1-4x-\frac{x^3}{3}+\frac{x^4}{12}-\frac{x^5}{60}+20\cdot O(x^6)=0$$
The first approximation $1-4x=0$ gives $\color{red}{x\approx 0.25}$
The second approximation $ 1-4x-\frac{x^3}{3}=0$ gives $\color{red}{x\approx 0.24872}$
And we can  continue but we see that the first approach is already good enough.
