Answer
$y=\left\{ -\dfrac{1}{4},\dfrac{2}{3} \right\}$
Work Step by Step
Factoring the given equation, $
12y^2-5y=2
,$ results to
\begin{array}{l}\require{cancel}
12y^2-5y-2=0
\\\\
(3y-2)(4y+1)=0
.\end{array}
Equating each factor to zero (Zero Product Principle), then the solutions to the equation, $
(3y-2)(4y+1)=0
,$ are
\begin{array}{l}\require{cancel}
3y-2=0
\\\\
3y=2
\\\\
y=\dfrac{2}{3}
,\\\\\text{OR}\\\\
4y+1=0
\\\\
4y=-1
\\\\
y=-\dfrac{1}{4}
.\end{array}
Hence, $
y=\left\{ -\dfrac{1}{4},\dfrac{2}{3} \right\}
.$
