$ 2 + \sqrt 3$ is root $ (x - 2)^2 = 3$ or $ x^2 - 4x + 1 = 0$ $ f(x) = (x + 4 + a)(x^2 - 4x + 1) + Ax - B$, were $ A = 4a + b + 15, B = a + 4 - c$. If $ A = 0,B = 0$ ($ b = - 4a - 15, - 6\le a\le - 2, c = a + 4$) number $ 2 + \sqrt 3$ is root of $ f(x) = 0$. Obviosly $ - 16\le c\le 24$. If $ B = 0,A\not = 0$, then $ |f(2 + \sqrt 3 )|\ge 2 + \sqrt 3.$ If $ A = 0,B\not = 0$, then $ |f(2 + \sqrt 3)|\ge 1$. If $ B\not = 0$, then $ |f(2 + \sqrt 3)|\ge min(|A_1(2 + \sqrt 3) - B|,|A_2(2 + \sqrt 3) - B|)$, were $ A_1 = [B(2 - \sqrt 3)],A_2 = A_1 + 1$. Because $ - 16\le B\le 24$, we get $ f(2 + \sqrt 3)|\ge ||A(2 + \sqrt 3)||, 1\le A\le 6$, were $ ||x|| = min(x - [x],[x] + 1 - x)$. Because $ ||x|| = ||\{x\}|| = ||\{x\} + j$, j - integer, we get $ |f(2 + \sqrt 3)|\ge min (||j\sqrt 3||, 1\le j\le 6)\ge ||4\sqrt 3|| = 0.0718$ if $ 2+\sqrt 3$ is not root.