To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$, where $x_{1}$ and $x_{2}$ are the solutions of the quadratic equation $ax^{2}+bx+c=0$.
$$5y^{2}-2y-3=0$$
All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. Substitute $5$ for $a$, $-2$ for $b$, and $-3$ for $c$ in the quadratic formula.
Solve the equation $y=\frac{2±8}{10}$ when $±$ is plus and when $±$ is minus.
$$y=1$$ $$y=-\frac{3}{5}$$
Rewrite the inequality by using the obtained solutions.
$$5\left(y-1\right)\left(y+\frac{3}{5}\right)>0$$
For the product to be positive, $y-1$ and $y+\frac{3}{5}$ have to be both negative or both positive. Consider the case when $y-1$ and $y+\frac{3}{5}$ are both negative.
$$y-1<0$$ $$y+\frac{3}{5}<0$$
The solution satisfying both inequalities is $y<-\frac{3}{5}$.
$$y<-\frac{3}{5}$$
Consider the case when $y-1$ and $y+\frac{3}{5}$ are both positive.
$$y+\frac{3}{5}>0$$ $$y-1>0$$
The solution satisfying both inequalities is $y>1$.
$$y>1$$
The final solution is the union of the obtained solutions.
