Answer
$(y+3)(20y-1)$
Work Step by Step
Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{
expression
}$
\begin{array}{l}\require{cancel}
20y^2+59y-3
\end{array} has $ac=
20(-3)=-60
$ and $b=
59
.$
The two numbers with a product of $c$ and a sum of $b$ are $\left\{
60,-1
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
20y^2+60y-1y-3
.\end{array}
Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to
\begin{array}{l}\require{cancel}
(20y^2+60y)-(y+3)
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
20y(y+3)-(y+3)
.\end{array}
Factoring the $GCF=
(y+3)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(y+3)(20y-1)
.\end{array}