To solve the equation (x-3)x (2y+1) = 7, we can start by expanding the expression:

(x-3)x (2y+1) = 7
(x^2 - 3x)(2y+1) = 7

Next, distribute the terms:

2xy^2 + xy - 6xy - 3x = 7

Combine like terms:

2xy^2 - 5xy - 3x = 7

Now, let's solve for x. We can factor out x from the equation:

x(2y^2 - 5y - 3) = 7

Now, set each factor equal to zero and solve for x:

x = 0 or 2y^2 - 5y - 3 = 7

For the second equation, let's solve for y:

2y^2 - 5y - 3 - 7 = 0
2y^2 - 5y - 10 = 0

Now, we can use the quadratic formula to solve for y:

y = (-(-5) ± √((-5)^2 - 4(2)(-10))) / (2(2))
y = (5 ± √(25 + 80)) / 4
y = (5 ± √105) / 4

So, the solutions for y are:

y = (5 + √105) / 4
y = (5 - √105) / 4

Therefore, the solutions for the equation (x-3)x (2y+1) = 7 are:

x = 0, y = (5 + √105) / 4
x = 0, y = (5 - √105) / 4
x = (5 + √105) / 4, y = (5 + √105) / 4
x = (5 + √105) / 4, y = (5 - √105) / 4