#### Answer

$\left\{ -\dfrac{1}{3},0,\dfrac{5}{2} \right\}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
Express the given equation, $
6x^3-13x^2-5x=0
,$ in factored form. Then use the Zero Product Property by equating each factor to zero. Finally, solve each of the resulting equations.
$\bf{\text{Solution Details:}}$
The $GCF$ of the constants of the terms $\{
6,-13,-5
\}$ is $
1
$ since it is the highest number that can divide all the given constants. The $GCF$ of the common variable/s is the variable/s with the lowest exponent. Hence, the $GCF$ of the common variable/s $\{
x^3,x^2,x
\}$ is $
x
.$ Hence, the entire expression has $GCF=
x
.$
Factoring the $GCF=
x
,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
x(6x^2-13x-5)=0
.\end{array}
Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $
6(-5)=-30
$ and the value of $b$ is $
-13
.$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{
2,-15
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
x(6x^2+2x-15x-5)=0
.\end{array}
Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to
\begin{array}{l}\require{cancel}
x[(6x^2+2x)-(15x+5)]=0
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
x[2x(3x+1)-5(3x+1)]=0
.\end{array}
Factoring the $GCF=
3x+1
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
x[(3x+1)(2x-5)]=0
\\\\
x(3x+1)(2x-5)=0
.\end{array}
Equating each factor zero (Zero Product Property), then
\begin{array}{l}\require{cancel}
x=0
\\\\\text{OR}\\\\
3x+1=0
\\\\\text{OR}\\\\
2x-5=0
.\end{array}
Solving each equation results to
\begin{array}{l}\require{cancel}
x=0
\\\\\text{OR}\\\\
3x+1=0
\\\\
3x=-1
\\\\
x=-\dfrac{1}{3}
\\\\\text{OR}\\\\
2x-5=0
\\\\
2x=5
\\\\
x=\dfrac{5}{2}
.\end{array}
Hence, the solutions are $
\left\{ -\dfrac{1}{3},0,\dfrac{5}{2} \right\}
.$