Answer:
[tex]\large\boxed{x=-8\pm5\sqrt2}[/tex]
Step-by-step explanation:
[tex](x-2)^2+12(x+2)-14=0\\\\\text{Substitute}\ (x+2)=t:\\\\t^2+12t-14=0\qquad\text{add 14 to both sides}\\\\t^2+2(t)(6)=14\qquad\text{add}\ 6^2=36\ \text{to both sides}\\\\\underbrace{t^2+2(t)(6)+6^2}=14+36\qquad\text{use}\ (a+b)^2=a^2+2ab+b^2\\\\(t+6)^2=50\Rightarrow t+6=\pm\sqrt{50}\qquad\text{subtract 6 from both sides}\\\\t=-6\pm\sqrt{25\cdot2}\\\\t=-6\pm5\sqrt2[/tex]
[tex]\text{we're going back to substitution}\\\\x+2=-6\pm5\sqrt2\qquad\text{subtract 2 from both sides}\\\\x=-8\pm5\sqrt2[/tex]
