#### Answer

$y=1\pm\sqrt{2}$

#### Work Step by Step

$\dfrac{1}{2}y^{2}=y+\dfrac{1}{2}$
Multiply the whole equation by $2$ to avoid working with fractions:
$2\Big(\dfrac{1}{2}y^{2}=y+\dfrac{1}{2}\Big)$
$y^{2}=2y+1$
Take all the terms on the right to the left side of the equation:
$y^{2}-2y-1=0$
Use the quadratic formula to solve this equation. The formula is $y=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. For this equation, $a=1$, $b=-2$ and $c=-1$
Substitute:
$y=\dfrac{-(-2)\pm\sqrt{(-2)^{2}-4(1)(-1)}}{2(1)}=\dfrac{2\pm\sqrt{4+4}}{2}=...$
$...=\dfrac{2\pm\sqrt{8}}{2}=\dfrac{2\pm2\sqrt{2}}{2}=1\pm\sqrt{2}$