#### Answer

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

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
Express the given equation, $
2x^3-9x^2-5x=0
,$ in factored form. Next, use the Zero-Factor Property by equating each factor to zero. Finally, solve each equation.
$\bf{\text{Solution Details:}}$
The $GCF$ of the constants of the terms $\{
2,-9,-5
\}$ is $
1
.$ 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 \left( 2x^2-9x-5 \right)=0
.\end{array}
To factor the trinomial expression above, find two numbers whose product is $ac$ and whose sum is $b$ in the quadratic expression $ax^2+bx+c.$ In the expression above the value of $ac$ is $x \left( 2x^2-9x-5 \right)=0
2(-5)=-10
$ and the value of $b$ is $
-9
.$
The possible pairs of integers whose product is $ac$ are
\begin{array}{l}\require{cancel}
\{1,-10\}, \{2,-5\},
\\
\{-1,10\}, \{-2,5\}
.\end{array}
Among these pairs, the one that gives a sum of $b$ is $\{
1,-10
\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
x \left( 2x^2+1x-10x-5 \right)=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 [(2x^2+1x)-(10x+5)]=0
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
x [x(2x+1)-5(2x+1)]=0
.\end{array}
Factoring the $GCF=
(2x+1)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
x [(2x+1)(x-5)]=0
\\\\
x(2x+1)(x-5)=0
.\end{array}
Equating each of the factors in the expression above to zero (Zero-Factor Theorem), then,
\begin{array}{l}\require{cancel}
x=0
\text{ OR }
2x+1=0
\text{ OR }
x-5=0
.\end{array}
Solving each of the equations above results to
\begin{array}{l}\require{cancel}
x=0
\\\\\text{OR}\\\\
2x+1=0
\\\\
2x=-1
\\\\
x=-\dfrac{1}{2}
\\\\\text{ OR }\\\\
x-5=0
\\\\
x=5
.\end{array}
Hence, the solutions are $
x=\left\{ -\dfrac{1}{2},0,5 \right\}
.$