Question 263557
I will use difference of two squares to get
{{{(4x-1)(4x+1) = 0}}}
set each part = 0 and solve
first,
{{{4x-1) = 0}}}
add 1 to get
{{{4x = 1}}}
divide by 4 to get
{{{x = 1/4}}}
second,
{{{4x+1) = 0}}}
subtract 1 to get
{{{4x = -1}}}
divide by 4 to get
{{{x = -1/4}}}