Answer :

[tex]2x^2+12x-14=0\ \ \ \ |divide\ both\ sides\ by\ 2\\\\x^2+6x-7=0\\\\x^2+2x\cdot3=7\ \ \ \ |add\ 3^2\ to\ both\ sides\\\\x^2+2x\cdot3+3^2=7+3^2\\\\(x+3)^2=7+9\\\\(x+3)^2=16\iff x+3=-\sqrt{16}\ or\ x+3=\sqrt{16}\\\\x+3=-4\ or\ x+3=4\\\\x=-4-3\ or\ x=4-3\\\\\boxed{x=-7\ or\ x=1}[/tex]
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kate200468
[tex] 2x^2 + 12x-14 = 0\\2(x^2+6x-7)=0\ /:2\\x^2+6x+9-16=0\\x^2+6x+9=16\\(x-3)^2=4^2\\x-3=4\ \ \ or\ \ \ x-3=-4\\x=7\ \ \ \ \ \ \ \ \ or\ \ \ x=-1\\\\Ans.\ x=7\ or\ x=-1[/tex]