Answer
$\left\{-\dfrac{11}{6},-\dfrac{19}{12}\right\}$
Work Step by Step
Let $z=
3x+5
,$ Then the given equation, $
8(3x+5)^2+2(3x+5)-1=0
$, is equivalent to
\begin{align*}
8z^2+2z-1&=0
.\end{align*}
Using factoring of trinomials, the equation above is equivalent to
\begin{align*}
(4z-1)(2z+1)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
4z-1=0 & 2z+1=0
\\
4z=1 & 2z=-1
\\\\
z=\dfrac{1}{4} & z=-\dfrac{1}{2}
.\end{array}
Since $z=
3x+5
,$ by back substitution, then
\begin{array}{l|r}\require{cancel}
3x+5=\dfrac{1}{4} & 3x+5=-\dfrac{1}{2}
\\\\
4(3x+5)=\left(\dfrac{1}{\cancel4}\right)\cancel4 & 2(3x+5)=-\left(\dfrac{1}{\cancel2}\right)\cancel2
\\\\
12x+20=1 & 6x+10=-1
\\
12x=1-20 & 6x=-1-10
\\
12x=-19 & 6x=-11
\\\\
x=-\dfrac{19}{12} & x=-\dfrac{11}{6}
.\end{array}
Hence, the solution set of the equation $
8(3x+5)^2+2(3x+5)-1=0
$ is $
\left\{-\dfrac{11}{6},-\dfrac{19}{12}\right\}
$.
