\begin{equation*}
\log_{10}{(x + 1)(x - 2)} = 1
\end{equation*}
Once the left-hand side is expressed as a single logarithm, we can rewrite the equation in exponential form as
\begin{equation*}
(x + 1)(x - 2) = 10^1
\end{equation*}
Simplifying the right side gives us a quadratic equation to solve.
\begin{align*}
x^2 - x - 2 \amp = 10 \amp \amp \blert{\text{Subtract 10 from both sides.}}\\
x^2 - x - 12 \amp = 0 \amp \amp \blert{\text{Factor the left side.}}\\
(x - 4)(x + 3) \amp= 0 \amp \amp \blert{\text{Apply the zero-factor principle.}}
\end{align*}
We find \(x = 4\) or \(x = -3\text{.}\) But we must check the original equation for extraneous solutions. The number \(-3\) is not a solution of the original equation, because neither \(\log_{10}{(x + 1)}\) nor \(\log_{10}{(x - 2)}\) is defined for \(x = -3\text{.}\) The apparent solution \(x=-3\) is extraneous, and the solution of the original equation is \(4\text{.}\)