Solve by Factoring 2y^2+10y=28

$2 y^{2} + 10 y = 28$

Step 1

Subtract $28$ from both sides of the equation.

$2 y^{2} + 10 y - 28 = 0$

Step 2

Factor $2$ out of $2 y^{2} + 10 y - 28$ .

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

$2 (y^{2}) + 10 y - 28 = 0$

Factor $2$ out of $10 y$ .

$2 (y^{2}) + 2 (5 y) - 28 = 0$

Factor $2$ out of $- 28$ .

$2 y^{2} + 2 (5 y) + 2 \cdot - 14 = 0$

Factor $2$ out of $2 y^{2} + 2 (5 y)$ .

$2 (y^{2} + 5 y) + 2 \cdot - 14 = 0$

Factor $2$ out of $2 (y^{2} + 5 y) + 2 \cdot - 14$ .

$2 (y^{2} + 5 y - 14) = 0$

Step 3

Factor.

Factor $y^{2} + 5 y - 14$ 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 $- 14$ and whose sum is $5$ .

$- 2, 7$

Write the factored form using these integers.

$2 ((y - 2) (y + 7)) = 0$

Remove unnecessary parentheses.

$2 (y - 2) (y + 7) = 0$

Step 4

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

$y - 2 = 0$

$y + 7 = 0$

Step 5

Set $y - 2$ equal to $0$ and solve for $y$ .

Set $y - 2$ equal to $0$ .

$y - 2 = 0$

Add $2$ to both sides of the equation.

$y = 2$

Step 6

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 7

The final solution is all the values that make $2 (y - 2) (y + 7) = 0$ true.

$y = 2, - 7$

Solve by Factoring 2y^2+10y=28

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