[tex]\it 16-8\sqrt3=12+4-8\sqrt3=(\sqrt{12})^2-8\sqrt3+2^2=(2\sqrt3-2)^2\\ \\ Egalitatea\ \ devine:\\ \\ \sqrt{(2\sqrt3-2)^2}=a+b\sqrt3 \Rightarrow |\underbrace{2\sqrt3-2}_{>0}|=a+b\sqrt3 \Rightarrow 2\sqrt3-2=a+b\sqrt3 \Rightarrow\\ \\ \Rightarrow a+b\sqrt3=-2+2\sqrt3 \Rightarrow a=-2,\ \ b=2[/tex]