Answer
$x=\frac{-1\pm\sqrt{5}}{2}$
Work Step by Step
We need to solve:
$x^{2}e^{x}+xe^{x}-e^{x}=0$
We factor:
$e^{x}(x^{2}+x-1)=0$
We have either:
$e^{x}=0$
$x=\ln 0$=no solution
Or:
$x^{2}+x-1=0$
$4x^{2}+4x-4=0$
$4x^{2}+4x+1-5=0$
$(2x+1)^2-5=0$
$(2x+1)^2=5$
$2x+1=\pm\sqrt{5}$
$2x=\pm\sqrt{5}-1$
$x=\frac{-1\pm\sqrt{5}}{2}$
(We could also have used the quadratic formula.)