Answer
$a=\left\{ -\dfrac{5}{2},\dfrac{4}{3} \right\}$
Work Step by Step
Using $(a+b)(c+d)=ac+ad+bc+bd$ or the Distributive Property, the given expression, $
(6a+1)(a+1)=21
,$ is equivalent to
\begin{array}{l}\require{cancel}
6a(a)+6a(1)+1(a)+1(1)=21
\\\\
6a^2+6a+a+1=21
\\\\
6a^2+6a+a+1-21=0
\\\\
6a^2+7a-20=0
.\end{array}
Factoring the above equation, $
6a^2+7a-20=0
,$ results to
\begin{array}{l}\require{cancel}
(2a+5)(3a-4)=0
.\end{array}
Equating each factor to zero (Zero Product Principle), then the solutions to the equation, $
(2a+5)(3a-4)=0
,$ are
\begin{array}{l}\require{cancel}
2a+5=0
\\\\
2a=-5
\\\\
a=-\dfrac{5}{2}
,\\\\\text{OR}\\\\
3a-4=0
\\\\
3a=4
\\\\
a=\dfrac{4}{3}
.\end{array}
Hence, $
a=\left\{ -\dfrac{5}{2},\dfrac{4}{3} \right\}
.$
