Answer
The solutions are $y=-\dfrac{1}{2}\pm\dfrac{\sqrt{5}}{4}$
Work Step by Step
$16y^{2}+16y-1=0$
Take $-1$ to the right side:
$16y^{2}+16y=1$
Take out common factor $16$ from the left side:
$16(y^{2}+y)=1$
Take $16$ to divide the right side:
$y^{2}+y=\dfrac{1}{16}$
Add $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation. For this particular case, $b=1$
$y^{2}+y+\Big(\dfrac{1}{2}\Big)^{2}=\dfrac{1}{16}+\Big(\dfrac{1}{2}\Big)^{2}$
$y^{2}+y+\dfrac{1}{4}=\dfrac{1}{16}+\dfrac{1}{4}$
$y^{2}+y+\dfrac{1}{4}=\dfrac{5}{16}$
Factor the left side of the equation, which is a perfect square trinomial:
$\Big(y+\dfrac{1}{2}\Big)^{2}=\dfrac{5}{16}$
Take the square root of both sides:
$\sqrt{\Big(y+\dfrac{1}{2}\Big)^{2}}=\pm\sqrt{\dfrac{5}{16}}$
$y+\dfrac{1}{2}=\pm\dfrac{\sqrt{5}}{4}$
Solve for $y$:
$y=-\dfrac{1}{2}\pm\dfrac{\sqrt{5}}{4}$
