To solve the equation (5x+1)(5x-3)-7x+1=0, we can start by expanding the expression (5x+1)(5x-3):

(5x+1)(5x-3) = 5x * 5x + 5x * (-3) + 1 * 5x + 1 * (-3)
= 25x^2 - 15x + 5x - 3
= 25x^2 - 10x - 3

Now, we can rewrite the equation as:

25x^2 - 10x - 3 - 7x + 1 = 0

Combining like terms:

25x^2 - 17x - 2 = 0

To solve this quadratic equation, we can use factoring, completing the square, or the quadratic formula. In this case, factoring is not straightforward, so we'll use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

For our equation, a = 25, b = -17, and c = -2. Plugging these values into the quadratic formula:

x = (-(-17) ± √((-17)^2 - 4 * 25 * (-2))) / (2 * 25)
x = (17 ± √(289 + 200)) / 50
x = (17 ± √489) / 50

Therefore, the solutions to the equation (5x+1)(5x-3)-7x+1=0 are:

x = (17 + √489) / 50
x = (17 - √489) / 50