Answer
$y=-3\pm\sqrt{\dfrac{15}{2}}$
Work Step by Step
$2y^{2}+12y+3=0$
Take the $3$ to the right side of the equation:
$2y^{2}+12y=-3$
Take out common factor $2$ from the left side of the equation:
$2(y^{2}+6y)=-3$
Add $\Big(\dfrac{b}{2}\Big)^{2}$ to the expression inside the parentheses and $2\Big(\dfrac{b}{2}\Big)^{2}$ to the right side of the equation. For this particular case, $b=6$
$2\Big[y^{2}+6y+\Big(\dfrac{6}{2}\Big)^{2}\Big]=-3+2\Big(\dfrac{6}{2}\Big)^{2}$
$2(y^{2}+6y+9)=-3+18$
$2(y^{2}+6y+9)=15$
Factor the expression inside the parentheses, which is a perfect square trinomial:
$2(y+3)^{2}=15$
Take the $2$ to divide the right side of the equation:
$(y+3)^{2}=\dfrac{15}{2}$
Take the square root of both sides of the equation:
$\sqrt{(y+3)^{2}}=\sqrt{\dfrac{15}{2}}$
$y+3=\pm\sqrt{\dfrac{15}{2}}$
Solve for $y$:
$y=-3\pm\sqrt{\dfrac{15}{2}}$
