We need to factor the quadratic expression $6y^2 + 7y - 3$.

1. Problem Description

We need to factor the quadratic expression $6y^2 + 7y - 3$.

2. Solution Steps

We are looking for two binomials of the form $(Ay + B)(Cy + D)$ such that

$AC = 6$, $AD + BC = 7$, and $BD = -3$.

We can try different combinations of factors for 6 and -

3. Possible factor pairs for 6 are (1, 6) and (2, 3).

Possible factor pairs for -3 are (1, -3) and (-1, 3).

Let's try $A = 2$ and $C = 3$, so we have $(2y + B)(3y + D)$.

We need $2D + 3B = 7$ and $BD = -3$.

If we try $B = 3$ and $D = -1$, we get $2(-1) + 3(3) = -2 + 9 = 7$. This works!

So the factorization is $(2y + 3)(3y - 1)$.

Let's check this:

$(2y + 3)(3y - 1) = 2y(3y) + 2y(-1) + 3(3y) + 3(-1) = 6y^2 - 2y + 9y - 3 = 6y^2 + 7y - 3$.

This is the correct expression.

3. Final Answer

$(2y + 3)(3y - 1)$