To solve the equation \(4x^2 - 1 = (2x + 1)(3x - 5)\), we need to expand the right-hand side and then simplify the equation.

First, let's expand the right-hand side:

\[
(2x + 1)(3x - 5)
\]

Using the distributive property (FOIL method):

\[
(2x + 1)(3x - 5) = 2x \cdot 3x + 2x \cdot (-5) + 1 \cdot 3x + 1 \cdot (-5)
\]

\[
= 6x^2 - 10x + 3x - 5
\]

\[
= 6x^2 - 7x - 5
\]

Now, we have the equation:

\[
4x^2 - 1 = 6x^2 - 7x - 5
\]

Next, let's move all terms to one side of the equation to set it to zero:

\[
4x^2 - 1 - 6x^2 + 7x + 5 = 0
\]

Combine like terms:

\[
(4x^2 - 6x^2) + 7x + (-1 + 5) = 0
\]

\[
-2x^2 + 7x + 4 = 0
\]

Now we have a quadratic equation:

\[
-2x^2 + 7x + 4 = 0
\]

To solve this quadratic equation, we can use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = -2\), \(b = 7\), and \(c = 4\).

First, calculate the discriminant:

\[
\Delta = b^2 - 4ac = 7^2 - 4(-2)(4)
\]

\[
= 49 + 32
\]

\[
= 81
\]

Since the discriminant is positive, we have two real solutions. Now, apply the quadratic formula:

\[
x = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-7 \pm \sqrt{81}}{2(-2)}
\]

\[
= \frac{-7 \pm 9}{-4}
\]

This gives us two solutions:

1. \(\frac{-7 + 9}{-4} = \frac{2}{-4} = -\frac{1}{2}\)

2. \(\frac{-7 - 9}{-4} = \frac{-16}{-4} = 4\)

So, the solutions to the equation \(4x^2 - 1 = (2x + 1)(3x - 5)\) are:

\[
x = -\frac{1}{2} \quad \text{and} \quad x = 4
\]