[tex](x+1)^2-4(x+1)+2=0\qquad\text{substitute}\ u=x+1\\\\u^2-4u+2=0\qquad\text{add 2 to both sides}\\\\u^2-4u+4=2\\\\u^2-2(u)(2)+2^2=2\qquad\text{use}\ (a-b)^2=a^2-2ab+b^2\\\\(u-2)^2=2\iff u-2=\pm\sqrt2\qquad\text{add 2 to both sides}\\\\u=2-\sqrt2\ \vee\ u=2+\sqrt2\\\\\text{Therefore}\\\\x+1=2-\sqrt2\ \vee\ x+1=2+\sqrt2\qquad\text{subtract 1 from both sides}\\\\\boxed{x=1-\sqrt2\ or\ x=1+\sqrt2}[/tex]
