$$y^{2}+9y+16-3y^{2}=-2$$
Combine $y^{2}$ and $-3y^{2}$ to get $-2y^{2}$.$$-2y^{2}+9y+16=-2$$
Subtract $16$ from both sides.$$-2y^{2}+9y=-2-16$$
Subtract $16$ from $-2$ to get $-18$.$$-2y^{2}+9y=-18$$
Divide both sides by $-2$.$$\frac{-2y^{2}+9y}{-2}=\frac{-18}{-2}$$
Dividing by $-2$ undoes the multiplication by $-2$.$$y^{2}+\frac{9}{-2}y=\frac{-18}{-2}$$
Divide $9$ by $-2$.$$y^{2}-\frac{9}{2}y=\frac{-18}{-2}$$
Divide $-18$ by $-2$.$$y^{2}-\frac{9}{2}y=9$$
Divide $-\frac{9}{2}$, the coefficient of the $x$ term, by $2$ to get $-\frac{9}{4}$. Then add the square of $-\frac{9}{4}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.$$y^{2}-\frac{9}{2}y+\left(-\frac{9}{4}\right)^{2}=9+\left(-\frac{9}{4}\right)^{2}$$
Square $-\frac{9}{4}$ by squaring both the numerator and the denominator of the fraction.$$y^{2}-\frac{9}{2}y+\frac{81}{16}=9+\frac{81}{16}$$
Add $9$ to $\frac{81}{16}$.$$y^{2}-\frac{9}{2}y+\frac{81}{16}=\frac{225}{16}$$
Factor $y^{2}-\frac{9}{2}y+\frac{81}{16}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.$$\left(y-\frac{9}{4}\right)^{2}=\frac{225}{16}$$
Take the square root of both sides of the equation.$$\sqrt{\left(y-\frac{9}{4}\right)^{2}}=\sqrt{\frac{225}{16}}$$
Simplify.$$y-\frac{9}{4}=\frac{15}{4}$$ $$y-\frac{9}{4}=-\frac{15}{4}$$
Add $\frac{9}{4}$ to both sides of the equation.$$y=6$$ $$y=-\frac{3}{2}$$