Answer
$x\approx 2.2639$
Work Step by Step
\begin{align*}
\ln{\left(x(x+1)\right)}&=2 &\text{Product Property of Logarithms.}\\\\
e^2&=x(x+1) &\text{Write the equation in exponential form.}\\\\
e^2&=x^2+x\\\\
0&=x^2+x-e^2 &\text{Subtract $e^2$ from both sides.}\\\\
\end{align*}
Recall:
The quadratic equation $a^2+bx+c=0$ can be solved using the Quadratic Formula:
$$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$$
With $a=1, b=1, $ and $c=-e^2$, solve using the Quadratic Formula to obtain:
\begin{align*}
x&=\dfrac{-1\pm \sqrt{1^2-4(1)(-e^2)}}{2(1)}\\\\
x&=\dfrac{-1\pm \sqrt{1+4e^2}}{2}\\\\
x&=\dfrac{-1\pm \sqrt{30.5562244}}{2}\\\\
x&=\dfrac{-1\pm 5.527768482}{2}\\\\
x_1&=\frac{-1-5.527768482}{2}\approx -3.2639\\\\
x_2=&=\frac{-1+5.527768482}{2}\approx 2.2639
\end{align*}
Since in $\ln{x}$, $x$ must be greater than $0$, then $x\approx-3.2639$ is not allowed.
Thus, the solution is $x\approx 2.2639$.
