Solve for: y ^ 2+9 y+16 = 3 y ^ 2-2

Expression: $$y ^ { 2 } + 9 y + 16 = 3 y ^ { 2 } - 2$$

Subtract $3y^{2}$ from both sides.


Combine $y^{2}$ and $-3y^{2}$ to get $-2y^{2}$.


Subtract $16$ from both sides.


Subtract $16$ from $-2$ to get $-18$.


Divide both sides by $-2$.


Dividing by $-2$ undoes the multiplication by $-2$.


Divide $9$ by $-2$.


Divide $-18$ by $-2$.


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.


Square $-\frac{9}{4}$ by squaring both the numerator and the denominator of the fraction.


Add $9$ to $\frac{81}{16}$.


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 $\left(x+\frac{b}{2}\right)^{2}$.


Take the square root of both sides of the equation.



$$y-\frac{9}{4}=\frac{15}{4}$$ $$y-\frac{9}{4}=-\frac{15}{4}$$

Add $\frac{9}{4}$ to both sides of the equation.

$$y=6$$ $$y=-\frac{3}{2}$$