Answer
$3y^{2}+\displaystyle \frac{3}{2}y-\frac{7}{4}-\frac{ \frac{7}{4}}{2y-1}$
Work Step by Step
$\begin{array}{ccccccccccc}
& & 3y^{2} & +\displaystyle \frac{3}{2}y& -\displaystyle \frac{7}{4} & \\
& &--&-- &--& \\
2y-1&) & 6y^{3} &+0 & -5y& +0 & & \\
& & 6y^{3} & -3y^{2} & & & \color{red}{\leftarrow \small{3y^{2}(2y-1) } } \\
& &--&-- & & & \color{red}{ \small{subtract}} \\
& & & 3y^{2} & -5y& +0 & \\
& & & 3y^{2} & -\displaystyle \frac{3}{2}y & & \color{red}{\leftarrow \small{\displaystyle \frac{3}{2}y(2y-1) } }\\
& & &--&-- & &\color{red}{ \small{subtract}} \\
& & & & -\displaystyle \frac{7}{2}y & +0 & \\
& & & & -\displaystyle \frac{7}{2}y & +\displaystyle \frac{7}{4} & \color{red}{\leftarrow \small{-\displaystyle \frac{7}{4}(2y-1) } }\\
& & & &-- & -- &\color{red}{ \small{subtract}} \\
& & & & & -\displaystyle \frac{7}{4} &
\end{array}$
Quotient = $3y^{2}+\displaystyle \frac{3}{2}y-\frac{7}{4}$
Remainder = $ -\displaystyle \frac{7}{4}$.
$\displaystyle \frac{dividend}{divisor}=quotient+\frac{remainder}{divisor}$
$\displaystyle \frac{ 6y^{3}-5y }{2y-1}$ = $3y^{2}+\displaystyle \frac{3}{2}y-\frac{7}{4}-\frac{ \frac{7}{4}}{2y-1}$
