Answer
$y=-1\pm\sqrt{\dfrac{7}{3}}$
Work Step by Step
$3y^{2}+6y-4=0$
Take the $4$ to the right side of the equation:
$3y^{2}+6y=4$
Take out common factor $3$ from the left side of the equation:
$3(y^{2}+2y)=4$
Add $\Big(\dfrac{b}{2}\Big)^{2}$ to the expression inside the parentheses and $3\Big(\dfrac{b}{2}\Big)^{2}$ to the right side of the equation. For this particular case, $b=2$
$3\Big[y^{2}+2y+\Big(\dfrac{2}{2}\Big)^{2}\Big]=4+3\Big(\dfrac{2}{2}\Big)^{2}$
$3(y^{2}+2y+1)=4+3$
$3(y^{2}+2y+1)=7$
Factor the expression inside the parentheses, which is a perfect square trinomial:
$3(y+1)^{2}=7$
Take the $3$ to divide the right side of the equation:
$(y+1)^{2}=\dfrac{7}{3}$
Take the square root of both sides of the equation:
$\sqrt{(y+1)^{2}}=\pm\sqrt{\dfrac{7}{3}}$
$y+1=\pm\sqrt{\dfrac{7}{3}}$
Solve for $y$:
$y=-1\pm\sqrt{\dfrac{7}{3}}$