Solve (x−1)2=12(x-1)^2 = 12(x−1)2=12.Ax=−1±23x = -1 \pm 2\sqrt{3}x=−1±23Bx=1±23x = 1 \pm 2\sqrt{3}x=1±23check_circleCx=1+23x = 1 + 2\sqrt{3}x=1+23Dx=1±12x = 1 \pm \sqrt{12}x=1±12 onlyExplanationx−1=±23x - 1 = \pm 2\sqrt{3}x−1=±23, so x=1±23x = 1 \pm 2\sqrt{3}x=1±23.