Solve for y (2y-1)(y+2)=y^2+16

$(2 y - 1) (y + 2) = y^{2} + 16$

Simplify $(2 y - 1) (y + 2)$ .

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Expand $(2 y - 1) (y + 2)$ using the FOIL Method.

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Apply the distributive property.

$2 y (y + 2) - 1 (y + 2) = y^{2} + 16$

Apply the distributive property.

$2 y \cdot y + 2 y \cdot 2 - 1 (y + 2) = y^{2} + 16$

Apply the distributive property.

$2 y \cdot y + 2 y \cdot 2 - 1 y - 1 \cdot 2 = y^{2} + 16$

Simplify and combine like terms.

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Simplify each term.

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Multiply $y$ by $y$ by adding the exponents.

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Move $y$ .

$2 (y \cdot y) + 2 y \cdot 2 - 1 y - 1 \cdot 2 = y^{2} + 16$

Multiply $y$ by $y$ .

$2 y^{2} + 2 y \cdot 2 - 1 y - 1 \cdot 2 = y^{2} + 16$

Multiply $2$ by $2$ .

$2 y^{2} + 4 y - 1 y - 1 \cdot 2 = y^{2} + 16$

Rewrite $- 1 y$ as $- y$ .

$2 y^{2} + 4 y - y - 1 \cdot 2 = y^{2} + 16$

Multiply $- 1$ by $2$ .

$2 y^{2} + 4 y - y - 2 = y^{2} + 16$

Subtract $y$ from $4 y$ .

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

Move all terms containing $y$ to the left side of the equation.

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Subtract $y^{2}$ from both sides of the equation.

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

Subtract $y^{2}$ from $2 y^{2}$ .

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

Move $16$ to the left side of the equation by subtracting it from both sides.

$y^{2} + 3 y - 2 - 16 = 0$

Subtract $16$ from $- 2$ .

$y^{2} + 3 y - 18 = 0$

Factor $y^{2} + 3 y - 18$ using the AC method.

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Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 18$ and whose sum is $3$ .

$- 3, 6$

Write the factored form using these integers.

$(y - 3) (y + 6) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$y - 3 = 0$

$y + 6 = 0$

Set the first factor equal to $0$ and solve.

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Set the first factor equal to $0$ .

$y - 3 = 0$

Add $3$ to both sides of the equation.

$y = 3$

Set the next factor equal to $0$ and solve.

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Set the next factor equal to $0$ .

$y + 6 = 0$

Subtract $6$ from both sides of the equation.

$y = - 6$

The final solution is all the values that make $(y - 3) (y + 6) = 0$ true.

$y = 3, - 6$

Solve for y (2y-1)(y+2)=y^2+16

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