Solve Using the Quadratic Formula 2y^2+12y+10=0

$2 y^{2} + 12 y + 10 = 0$

Step 1

Use the quadratic formula to find the solutions.

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

Step 2

Substitute the values $a = 2$ , $b = 12$ , and $c = 10$ into the quadratic formula and solve for $y$ .

$\frac{- 12 \pm \sqrt{12^{2} - 4 \cdot (2 \cdot 10)}}{2 \cdot 2}$

Step 3

Simplify.

Simplify the numerator.

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

$y = \frac{- 12 \pm \sqrt{144 - 4 \cdot 2 \cdot 10}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot 10$ .

Multiply $- 4$ by $2$ .

$y = \frac{- 12 \pm \sqrt{144 - 8 \cdot 10}}{2 \cdot 2}$

Multiply $- 8$ by $10$ .

$y = \frac{- 12 \pm \sqrt{144 - 80}}{2 \cdot 2}$

Subtract $80$ from $144$ .

$y = \frac{- 12 \pm \sqrt{64}}{2 \cdot 2}$

Rewrite $64$ as $8^{2}$ .

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

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

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

Multiply $2$ by $2$ .

$y = \frac{- 12 \pm 8}{4}$

Simplify $\frac{- 12 \pm 8}{4}$ .

$y = - 3 \pm 2$

Step 4

The final answer is the combination of both solutions.

$y = - 1, - 5$