2b^2+4b–2=0 what is the answer?
\(2b^2+4b-2=0\\ 2(b^2+2b-1)=0\\ 2(b^2+2b+1-2)=0\\ 2((b+1)^2-2)=0\\ 2(b+1)^2-4=0\\\ 2(b+1)^2=4\\\ (b+1)^2=2\\ b+1=-\sqrt2 \vee b+1=\sqrt2\\ b=-1-\sqrt2 \vee b=-1+\sqrt2\)
