Answer
The solutions are $y=-2$ and $y=\pm\dfrac{1}{2}$
Work Step by Step
$4y^{3}-2=y-8y^{2}$
Take all terms to the left side of the equation:
$4y^{3}+8y^{2}-y-2=0$
Group the first two terms together and the last two terms together:
$(4y^{3}+8y^{2})-(y+2)=0$
Take out common factor $4y^{2}$ from the first group:
$4y^{2}(y+2)-(y+2)=0$
Factor out $(y+2)$:
$(y+2)(4y^{2}-1)=0$
Set both factors equal to $0$ and solve each individual equation for $y$:
$y+2=0$
$y=-2$
$4y^{2}-1=0$
$4y^{2}=1$
$y^{2}=\dfrac{1}{4}$
$\sqrt{y^{2}}=\pm\sqrt{\dfrac{1}{4}}$
$y=\pm\dfrac{1}{2}$
The solutions are $y=-2$ and $y=\pm\dfrac{1}{2}$
