Answer
$\{-\frac{1}{4},\frac{3}{4}\}$
Work Step by Step
Add $\frac{3}{16}$ to both sides.
$x^2-\dfrac{1}{2}x-\dfrac{3}{16}+\dfrac{3}{16}=0+\dfrac{3}{16}$
Simplify.
$x^2-\dfrac{1}{2}x=\dfrac{3}{16}$
Complete the square by adding $\left(\dfrac{\frac{-1}{2}}{2}\right)^2=\frac{1}{16}$ to both sides.
$x^2-\dfrac{1}{2}x+\dfrac{1}{16}=\dfrac{3}{16}+\dfrac{1}{16}$
Simplify.
$x^2-\dfrac{1}{2}x+\dfrac{1}{16}=\dfrac{3+1}{16}\\$
$\\x^2-\dfrac{1}{2}x+\dfrac{1}{16}=\dfrac{4}{16}\\$
$\\x^2-\dfrac{1}{2}x+\dfrac{1}{16}=\dfrac{1}{4}$
Factor on the left side.
$\left(x-\dfrac{1}{4}\right)^2=\dfrac{1}{4}$
Take the square root of both sides:
$x-\dfrac{1}{4}=\pm \sqrt {\dfrac{1}{4}}$
$x-\dfrac{1}{4}=\pm \dfrac{1}{2}$
Split the expression to obtain:
$x-\dfrac{1}{4}=\dfrac{1}{2}\quad $ or $\quad x-\dfrac{1}{4}=-\dfrac{1}{2}$
Add $\frac{1}{4}$ to both sides.
$x-\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2}+\dfrac{1}{4}\quad$ or $\quad x-\dfrac{1}{4}+\dfrac{1}{4}=-\dfrac{1}{2}+\dfrac{1}{4}$
Simplify.
$x=\dfrac{3}{4}\quad $ or $\quad x=-\dfrac{1}{4}$
Hence, the solution set is $\{-\frac{1}{4},\frac{3}{4}\}$.
