Solve Using the Quadratic Formula 2y(y+6)=14

$2 y (y + 6) = 14$

Step 1

Move all terms to the left side of the equation and simplify.

Simplify the left side.

Simplify $2 y (y + 6)$ .

Apply the distributive property.

$2 y \cdot y + 2 y \cdot 6 = 14$

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

Move $y$ .

$2 (y \cdot y) + 2 y \cdot 6 = 14$

Multiply $y$ by $y$ .

$2 y^{2} + 2 y \cdot 6 = 14$

Multiply $6$ by $2$ .

$2 y^{2} + 12 y = 14$

Subtract $14$ from both sides of the equation.

$2 y^{2} + 12 y - 14 = 0$

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3

Substitute the values $a = 2$ , $b = 12$ , and $c = - 14$ into the quadratic formula and solve for $y$ .

$\frac{- 12 \pm \sqrt{12^{2} - 4 \cdot (2 \cdot - 14)}}{2 \cdot 2}$

Step 4

Simplify.

Simplify the numerator.

Raise $12$ to the power of $2$ .

$y = \frac{- 12 \pm \sqrt{144 - 4 \cdot 2 \cdot - 14}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot - 14$ .

Multiply $- 4$ by $2$ .

$y = \frac{- 12 \pm \sqrt{144 - 8 \cdot - 14}}{2 \cdot 2}$

Multiply $- 8$ by $- 14$ .

$y = \frac{- 12 \pm \sqrt{144 + 112}}{2 \cdot 2}$

Add $144$ and $112$ .

$y = \frac{- 12 \pm \sqrt{256}}{2 \cdot 2}$

Rewrite $256$ as $16^{2}$ .

$y = \frac{- 12 \pm \sqrt{16^{2}}}{2 \cdot 2}$

Pull terms out from under the radical, assuming positive real numbers.

$y = \frac{- 12 \pm 16}{2 \cdot 2}$

Multiply $2$ by $2$ .

$y = \frac{- 12 \pm 16}{4}$

Simplify $\frac{- 12 \pm 16}{4}$ .

$y = - 3 \pm 4$

Step 5

The final answer is the combination of both solutions.

$y = 1, - 7$