Answer:
x = -7 or x = 1
Step-by-step explanation:
[tex](a+b)^2=a^2+2ab+b^2\qquad(*)\\\\\\2x^2+12x-14=0\qquad\text{divide both sides by 2}\\\\x^2+6x-7=0\qquad\text{add 7 to both sides}\\\\x^2+2(x)(3)=7\qquad\text{add}\ 3^2=9\ \text{to both sides}\\\\\underbrace{x^2+2(x)(3)+3^2}_{(*)}=7+9\\\\(x+3)^2=16\Rightarrow x+3=\pm\sqrt{16}\\\\x+3=-4\ or\ x+3=4\qquad\text{subtract 3 from both sides}\\\\x=-7\ or\ x=1[/tex]
