#### Answer

$x=\left\{ -\dfrac{4}{3}, 5 \right\}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given equation, $
\log_4[(3x+1)(x-4)]=2
,$ convert to exponential form. Then use concepts on solving quadratic equations. Do checking of solutions with the original equation.
$\bf{\text{Solution Details:}}$
Since $\log_by=x$ is equivalent to $y=b^x$, the equation above, in exponential form, is equivalent to
\begin{array}{l}\require{cancel}
(3x+1)(x-4)=4^2
\\\\
(3x+1)(x-4)=16
.\end{array}
Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to\begin{array}{l}\require{cancel}
3x(x)+3x(-4)+1(x)+1(-4)=16
\\\\
3x^2-12x+x-4=16
\\\\
3x^2+(-12x+x)+(-4-16)=0
\\\\
3x^2-11x-20=0
.\end{array}
Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the factored form of the equation above is to
\begin{array}{l}\require{cancel}
(3x+4)(x-5)=0
.\end{array}
Equating each factor to zero (Zero Product Property), the solutions are
\begin{array}{l}\require{cancel}
3x+4=0
\\\\\text{OR}\\\\
x-5=0
.\end{array}
Solving each equation results to
\begin{array}{l}\require{cancel}
3x+4=0
\\\\
3x=-4
\\\\
x=-\dfrac{4}{3}
\\\\\text{OR}\\\\
x-5=0
\\\\
x=5
.\end{array}
Upon checking, both $
x=\left\{ -\dfrac{4}{3}, 5 \right\}
$ satisfy the original equation.