Solve for y -y+28+y^2=2y+2y^2

$- y + 28 + y^{2} = 2 y + 2 y^{2}$

Step 1

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

Subtract $2 y$ from both sides of the equation.

$- y + 28 + y^{2} - 2 y = 2 y^{2}$

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

$- y + 28 + y^{2} - 2 y - 2 y^{2} = 0$

Subtract $2 y$ from $- y$ .

$- 3 y + 28 + y^{2} - 2 y^{2} = 0$

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

$- 3 y + 28 - y^{2} = 0$

Step 2

Factor the left side of the equation.

Factor $- 1$ out of $- 3 y + 28 - y^{2}$ .

Reorder the expression.

Move $28$ .

$- 3 y - y^{2} + 28 = 0$

Reorder $- 3 y$ and $- y^{2}$ .

$- y^{2} - 3 y + 28 = 0$

Factor $- 1$ out of $- y^{2}$ .

$- (y^{2}) - 3 y + 28 = 0$

Factor $- 1$ out of $- 3 y$ .

$- (y^{2}) - (3 y) + 28 = 0$

Rewrite $28$ as $- 1 (- 28)$ .

$- (y^{2}) - (3 y) - 1 \cdot - 28 = 0$

Factor $- 1$ out of $- (y^{2}) - (3 y)$ .

$- (y^{2} + 3 y) - 1 \cdot - 28 = 0$

Factor $- 1$ out of $- (y^{2} + 3 y) - 1 (- 28)$ .

$- (y^{2} + 3 y - 28) = 0$

Factor.

Factor $y^{2} + 3 y - 28$ using the AC method.

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 $- 28$ and whose sum is $3$ .

$- 4, 7$

Write the factored form using these integers.

$- ((y - 4) (y + 7)) = 0$

Remove unnecessary parentheses.

$- (y - 4) (y + 7) = 0$

Step 3

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

$y - 4 = 0$

$y + 7 = 0$

Step 4

Set $y - 4$ equal to $0$ and solve for $y$ .

Set $y - 4$ equal to $0$ .

$y - 4 = 0$

Add $4$ to both sides of the equation.

$y = 4$

Step 5

Set $y + 7$ equal to $0$ and solve for $y$ .

Set $y + 7$ equal to $0$ .

$y + 7 = 0$

Subtract $7$ from both sides of the equation.

$y = - 7$

Step 6

The final solution is all the values that make $- (y - 4) (y + 7) = 0$ true.

$y = 4, - 7$

Solve for y -y+28+y^2=2y+2y^2

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