Solve for y 2y^2-28-7y=y^2-4y

$2 y^{2} - 28 - 7 y = y^{2} - 4 y$

Step 1

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

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

$2 y^{2} - 28 - 7 y - y^{2} = - 4 y$

Add $4 y$ to both sides of the equation.

$2 y^{2} - 28 - 7 y - y^{2} + 4 y = 0$

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

$y^{2} - 28 - 7 y + 4 y = 0$

Add $- 7 y$ and $4 y$ .

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

Step 2

Factor $y^{2} - 28 - 3 y$ 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 $- 3$ .

$- 7, 4$

Write the factored form using these integers.

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

Step 3

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

$y - 7 = 0$

$y + 4 = 0$

Step 4

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

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

$y - 7 = 0$

Add $7$ to both sides of the equation.

$y = 7$

Step 5

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

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

$y + 4 = 0$

Subtract $4$ from both sides of the equation.

$y = - 4$

Step 6

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

$y = 7, - 4$

Solve for y 2y^2-28-7y=y^2-4y

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