[tex]\displaystyle \\
A = \{ x \in R | (x-1)^2 = 3\} \\
\texttt{Rezolvam ecuatia: } \\
(x-1)^2 = 3 \\
x^2 -2x + 1 = 3 \\
x^2 -2x + 1 - 3 = 0 \\
x^2 -2x -2 = 0 \\ \\
x_{12}= \frac{-b \pm \sqrt{b^2 - 4ac} }{2a} = \frac{2 \pm \sqrt{4 +8} }{2} = \frac{2 \pm 2\sqrt{3} }{2} = 1 \pm \sqrt{3} \\ \\
x_1 = \boxed{1 + \sqrt{3} }\\
x_2 = \boxed{1 - \sqrt{3} }\\ \\
A = \{ 1 + \sqrt{3}, ~1 - \sqrt{3} \} \\ \\
\boxed{S = 1 + \sqrt{3} + 1 - \sqrt{3} = 1+1 = 2 }[/tex]