Expanding both sides of the equation:

(y-3)(5+2y) - (5-2y)^2 = 0

= 5y + 2y^2 - 15 - 6y - (25 - 20y + 4y^2)
= 2y^2 + 5y - 15 - 6y - 25 + 20y - 4y^2
= 2y^2 - 4y^2 + 5y - 6y + 20y - 15 - 25
= -2y^2 + 19y - 40

Setting this expression equal to zero to solve for y:

-2y^2 + 19y - 40 = 0

This is a quadratic equation, which can be solved using factoring, completing the square, or the quadratic formula. Factoring is a common method for this type of equation.

To factor the equation, we look for two numbers that multiply to -80 and add up to 19. These numbers are 20 and -4:

-2y^2 + 20y - 4y - 40 = 0
-2y(y - 10) - 4(y - 10) = 0
(y - 10)(-2y - 4) = 0
y = 10 or y = -2

So the solutions to the equation are y = 10 or y = -2.