Question 57766
{{{2w(4w+1)=1}}}
{{{8w^2+2w=1}}}
{{{8w^2+2w-1=1-1}}}
{{{8w^2+2w-1=0}}}
{{{8w^2+4w-2w-1=0}}}  I'm factoring by grouping,ac,or arch method. Use whatever method you're most comfortable.
{{{(8w^2+4w)+(-2w-1)=0}}}
{{{4w(2w+1)-1(2w+1)=0}}}
{{{(4w-1)(2w+1)=0}}}  Set each parenthesis = to 0.
4w-1=0  and 2w+1=0
4w=1  and 2w=-1
w=1/4 and w=-1/2
Happy Calculating!!!