To solve this equation, we will first expand and simplify the left side:

(х+4)(х-2) + (2-х)(2х-3)
= х^2 - 2х + 4х - 8 + 4х - 6х + 3
= х^2 - 2х + 4х - 8 + 4х - 6х + 3
= х^2 - 6х - 5

Now we set this expression equal to 0 and solve for x:

х^2 - 6х - 5 = 0

To factor this quadratic equation, we look for two numbers that multiply to -5 and add up to -6. These numbers are -5 and 1:

х^2 - 6х - 5 = (x - 5)(x + 1)

Setting each factor to zero:

x - 5 = 0
x = 5

x + 1 = 0
x = -1

Therefore, the solutions to the equation (х+4)(х-2) + (2-х)(2х-3) = 0 are x = 5 and x = -1.