Answer
$x=-\dfrac{1}{2}\pm\dfrac{\sqrt{5}}{2}$
Work Step by Step
$x^{2}e^{x}+xe^{x}-e^{x}=0$
Take out common factor $e^{x}$:
$e^{x}(x^{2}+x-1)=0$
Set both factors equal to $0$ and solve each individual equation:
$e^{x}=0$
No value of $x$ makes this first equation true. It has no solution. Move on to the second one:
$x^{2}+x-1=0$
Use the quadratic formula to solve this second equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. For this equation, $a=1$, $b=1$ and $c=-1$.
Substitute the known values into the formula and simplify:
$x=\dfrac{-1\pm\sqrt{1^{2}-4(1)(-1)}}{2(1)}=\dfrac{-1\pm\sqrt{1+4}}{2}=\dfrac{-1\pm\sqrt{5}}{2}=...$
$...=-\dfrac{1}{2}\pm\dfrac{\sqrt{5}}{2}$
The solutions are:
$x=-\dfrac{1}{2}\pm\dfrac{\sqrt{5}}{2}$
