College Algebra, Chapter 5, 5.4, Section 5.4, Problem 36

Solve the equation $x^2 e^x + x e^x - e^x = 0$

$
\begin{equation}
\begin{aligned}
x^2 e^x + x e^x - e^x &= 0\\
\\
e^x(x^2+x-1) &= 0 && \text{Factor out common terms}\\
\\
x^2 + x - 1 &= 0 && \text{Divide by $e^x$ (because $e^x \neq 0$)}
\end{aligned}
\end{equation}
$

Solve for $x$, using quadratic formula. We get

$
\begin{equation}
\begin{aligned}
x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\
\\
x &= \frac{-1\pm \sqrt{(1)^2 - 4(1)(-1)}}{2(1)}\\
\\
x &= \frac{-1\pm\sqrt{1+4}}{2}\\
\\
x &= \frac{-1\pm \sqrt{5}}{2}
\end{aligned}
\end{equation}
$

So the solution of the given equation are $\displaystyle x = \frac{-1+ \sqrt{5}}{2} \text{ and } x = \frac{-1-\sqrt{5}}{2}$