Answer
$y=\dfrac{-2\pm\sqrt{5}}{4}$
Work Step by Step
Using the completing the square method, the solutions of the given quadratic equation, $
16y^2+16y=1
,$ is
\begin{array}{l}\require{cancel}
\dfrac{16y^2+16y}{16}=\dfrac{1}{16}
\\\\
y^2+y=\dfrac{1}{16}
\\\\
y^2+y+\left( \dfrac{1}{2} \right)^2=\dfrac{1}{16}+\left( \dfrac{1}{2} \right)^2
\\\\
y^2+y+\dfrac{1}{4}=\dfrac{1}{16}+\dfrac{1}{4}
\\\\
\left( y+\dfrac{1}{2} \right)^2=\dfrac{1}{16}+\dfrac{4}{16}
\\\\
\left( y+\dfrac{1}{2} \right)^2=\dfrac{5}{16}
\\\\
y+\dfrac{1}{2}=\pm\sqrt{\dfrac{5}{16}}
\\\\
y+\dfrac{1}{2}=\pm\dfrac{\sqrt{5}}{\sqrt{16}}
\\\\
y+\dfrac{1}{2}=\pm\dfrac{\sqrt{5}}{4}
\\\\
y=-\dfrac{1}{2}\pm\dfrac{\sqrt{5}}{4}
\\\\
y=-\dfrac{2}{4}\pm\dfrac{\sqrt{5}}{4}
\\\\
y=\dfrac{-2\pm\sqrt{5}}{4}
.\end{array}
