How do you solve $2y=3{{y}^{2}}$?

Hint: We have been given a quadratic equation of $y$ as $2y=3{{y}^{2}}$. We use the quadratic formula to solve the value of the $y$. we have the solution in the form of $y=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ for general equation of $a{{y}^{2}}+by+c=0$. We put the values and find the solution.

Complete step by step answer:
We know for a general equation of quadratic $a{{y}^{2}}+by+c=0$, the value of the roots of $y$ will be $y=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. This is the quadratic equation solving method. The root part $\sqrt{{{b}^{2}}-4ac}$ of $y=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ is called the discriminant of the equation.
In the given equation we have $2y=3{{y}^{2}}$. Simplified form is $3{{y}^{2}}-2y=0$
The values of a, b, c is $3,-2,0$ respectively.
We put the values and get $y$ as \[y=\dfrac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\times 0\times 3}}{2\times 3}=\dfrac{2\pm \sqrt{4}}{6}=\dfrac{2\pm 2}{6}=0,\dfrac{2}{3}\]
The roots of the equation are real numbers.

So, values of $y$ are $y=0,\dfrac{2}{3}$.

Note: We need to find the solution of the given equation $2y=3{{y}^{2}}$.
The simplified form is $3{{y}^{2}}-2y=0$.
First, we try to take a common number or variable out of the terms $3{{y}^{2}}$ and $-2y$.
The only thing that can be taken out is $y$.
So, $3{{y}^{2}}-2y=y\left( 3y-2 \right)=0$.
The multiplication of two terms gives 0. This gives that at least one of the terms has to be zero.
We get the values of $y$ as either $y=0$ or $\left( 3y-2 \right)=0$.
This gives $y=0,\dfrac{2}{3}$.
The given quadratic equation has 2 solutions and they are $y=0,\dfrac{2}{3}$.