Answer
$\displaystyle \{-\frac{1}{4},\frac{3}{4}\}$
Work Step by Step
$x^{2}-\displaystyle \frac{1}{2}x-\frac{3}{16}=0$
multiply with the LCD, 16
$16x^{2}-8x-3=0$
The first two terms can be recognized as
$(4x)^{2}-2\cdot(4x)\cdot(1)$
The third term missing for a perfect square is $(1)^{2}$, so we rewrite the equation
$((4x)^{2}-2\cdot(4x)\cdot(1)+1^{2})-1^{2}-3^{2}=0$
$(4x-1)^{2}-4=0$
Now, with $4=2^{2}$, the LHS is a difference of squares, which factors to
$[(4x-1)+2][(4x-1)-2]=0$
$(4x+1)(4x-3)=0$
so either $ 4x+1=0\quad$or$\quad 4x-3=0$
$x=-\displaystyle \frac{1}{4}\quad $or$\displaystyle \quad x=\frac{3}{4}.$
Solution set = $\displaystyle \{-\frac{1}{4},\frac{3}{4}\}$
