# 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.

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

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.

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

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