Points of intersection between $y=e^{-x^{2}}$ and $y=x^{2}-1$ If I want to find the area enclosed by the curves 
$$y=e^{-x^{2}} \quad \text{and} \quad y=x^{2}-1$$
is there a way I can solve for $x$ here to find the points of intersection?
$$e^{-x^{2}}=x^{2}-1$$
or is graphing the only way I can find these points? Because I have tried but I am stuck.
 A: The points of intersection satisfy, of course, $e^{-x^2}= x^2- 1$. Let $u= x^2$ so the equation becomes $e^{-u}= u- 1$. Let z= u- 1 so u= z+ 1 and the equation becomes $e^{-z-1}= e^{-1}e^{-z}= z$.  Multiply both sides by $e^z$ to get $ze^{z}= e^{-1}$.  The solution to that is $z= W\left(e^{-1}\right)$ where "W" is "Lambert's W function",  (https://en.wikipedia.org/wiki/Lambert_W_function) which is defined as the inverse function to $f(x)= xe^x$.  
So $x= \pm\sqrt{u}= \pm\sqrt{z+ 1}= \pm\sqrt{W\left(e^{-1}\right)+ 1}$.
A: Hint. Use the Lambert W function that is the inverse of $f(x)=xe^x$.
A: This is not solvable exactly, but the solution can be approximated via iteration.
We find $g : f(x)=0\iff g(x)=x$, then $$f(\lim_{n\to\infty} x_n)=0 \text{ where }x_{n+1}=g(x_n)$$
Here: $e^{-x^2}=x^2-1\to $
$$x=\pm\sqrt{e^{-x^2}+1}\tag 1$$
$$x=\pm\sqrt{\ln\bigg(\frac{1}{x^2-1}\bigg)}\tag2$$
Use these two to show the solutions are $\pm1.130692064_C$ 
A: Without graphing.
If you do not want to use Lambert function, consider that you look for the zero(s) of function
$$f(x)=e^{-x}-x+1$$ The first derivatives are
$$f'(x)=-e^{-x}-1 < 0 \qquad \forall x  \qquad \text{and}\qquad f''(x)=e^{-x} > 0 \qquad \forall x$$ So, there is only one root to this equation and a numerical method can be used.
Newton method being the simplest, start from a guess $x_0$ which will be updated according to
$$x_{k+1}=x_k-\frac{f(x_k)}{f'(x_k)}$$ which, for the present case will be
$$x_{k+1}=1+\frac{x_k}{1+e^{x_k}}$$ Since the second derivative is positive, in order to avoid an overshoot of the solution, select $x_0$ such that $f(x_0) >0$ (this is by Darboux theorem).
Being lazy, let us use $x_0=0$ $(f(0)=2)$. The iterates would be
$$\left(
\begin{array}{cc}
 k & x_k \\
 0 & 0.000000000 \\
 1 & 1.000000000 \\
 2 & 1.268941421 \\
 3 & 1.278454624 \\
 4 & 1.278464543
\end{array}
\right)$$
Then, back to $x^2$, the roots are $\pm \sqrt{1.278464543}=\pm 1.130692064$.
