To solve the equation \((6x - 2)^2 + (5x + 2)^2 - 4(3x - 1)(5x + 2) = 0\), let's start by expanding each term.

First, expand \((6x - 2)^2\):
\[
(6x - 2)^2 = (6x)^2 - 2 \cdot 6x \cdot 2 + 2^2 = 36x^2 - 24x + 4
\]

Next, expand \((5x + 2)^2\):
\[
(5x + 2)^2 = (5x)^2 + 2 \cdot 5x \cdot 2 + 2^2 = 25x^2 + 20x + 4
\]

Now, expand \(4(3x - 1)(5x + 2)\):
\[
4(3x - 1)(5x + 2) = 4 \left( 3x \cdot 5x + 3x \cdot 2 - 1 \cdot 5x - 1 \cdot 2 \right) = 4 \left( 15x^2 + 6x - 5x - 2 \right) = 4 \left( 15x^2 + x - 2 \right) = 60x^2 + 4x - 8
\]

Now, substitute these expanded forms back into the original equation:
\[
36x^2 - 24x + 4 + 25x^2 + 20x + 4 - (60x^2 + 4x - 8) = 0
\]

Combine like terms:
\[
(36x^2 + 25x^2 - 60x^2) + (-24x + 20x - 4x) + (4 + 4 + 8) = 0
\]
\[
(61x^2 - 60x^2) + (-24x + 20x - 4x) + 16 = 0
\]
\[
x^2 - 8x + 16 = 0
\]

This is a quadratic equation. We can solve it using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -8\), and \(c = 16\).

First, calculate the discriminant:
\[
b^2 - 4ac = (-8)^2 - 4 \cdot 1 \cdot 16 = 64 - 64 = 0
\]

Since the discriminant is 0, there is exactly one real solution:
\[
x = \frac{-(-8) \pm \sqrt{0}}{2 \cdot 1} = \frac{8}{2} = 4
\]

Thus, the solution to the equation is:
\[
x = 4
\]