Решите уравнение f(x+2)=f(x-2)+4, если f(x)=3+2x+x^2

Решите задачу:

$f(x) = 3 + 2x + {x}^{2} \\ f(x + 2) = 3 + 2(x + 2) + {(x + 2)}^{2} = 3 + 2x + 4 + {x}^{2} + 4x + 4 = {x}^{2} + 6x + 11 \\ f(x - 2) = 3 + 2( x - 2) + {(x - 2)}^{2} = 3 + 2x - 4 + {x}^{2} - 4x + 4 = {x}^{2} - 2x + 3 \\ {x}^{2} + 6x + 11 = {x}^{2} - 2x + 3 + 4 \\ {x}^{2} - {x}^{2} + 6x + 2x + 11 - 3 - 4 = 0 \\ 8x + 4 = 0 \\ 8x = - 4 \\ x = - \frac{1}{2}$