[tex]\displaystyle 4 \cdot \left( \frac{2+ \sqrt{3} }{2} \right)^2-8 \cdot \left( \frac{2+ \sqrt{3} }{2} \right)+1=4 \cdot \frac{\left(2+ \sqrt{3} \right)^2}{2^2} - \frac{\not8 \left(2+ \sqrt{3} \right)}{\not2}+1= \\ \\ =\not4 \cdot \frac{2^2+2 \cdot 2 \cdot \sqrt{3}+\left( \sqrt{3}\right)^2 }{\not4} - 4\left(2+ \sqrt{3} \right)+1= \\ \\ =4+4 \sqrt{3} +3-8-4 \sqrt{3} +1=0[/tex]