To factor the quadratic expression \(8y^2 + 10y - 3\), we can use the method of factoring by grouping or applying the quadratic formula to find the roots.
The standard form of the quadratic is \(ax^2 + bx + c\), where:
- \(a = 8\)
- \(b = 10\)
- \(c = -3\)
We first calculate the product \(ac\):
\[ ac = 8 \times -3 = -24 \]
We need to find two numbers that multiply to \(-24\) (the value of \(ac\)) and add up to \(10\) (the value of \(b\)). The numbers that satisfy this condition are \(12\) and \(-2\) since:
\[ 12 \times -2 = -24 \quad \text{and} \quad 12 + (-2) = 10 \]
Next, we can rewrite the middle term using these two numbers:
\[ 8y^2 + 12y - 2y - 3 \]
Now we group the terms:
\[ (8y^2 + 12y) + (-2y - 3) \]
Factor out the common factors in each group:
\[ 4y(2y + 3) - 1(2y + 3) \]
Now we can factor out the common binomial factor \((2y + 3)\):
\[ (2y + 3)(4y - 1) \]
Thus, the factored form of the quadratic expression \(8y^2 + 10y - 3\) is:
\[ \boxed{(2y + 3)(4y - 1)} \]