Solve Using the Quadratic Formula y^2+12y+20=0

$y^{2} + 12 y + 20 = 0$

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 $a = 1$ , $b = 12$ , and $c = 20$ into the quadratic formula and solve for $y$ .

$\frac{- 12 \pm \sqrt{12^{2} - 4 \cdot (1 \cdot 20)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

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

$y = \frac{- 12 \pm \sqrt{144 - 4 \cdot 1 \cdot 20}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 20$ .

Multiply $- 4$ by $1$ .

$y = \frac{- 12 \pm \sqrt{144 - 4 \cdot 20}}{2 \cdot 1}$

Multiply $- 4$ by $20$ .

$y = \frac{- 12 \pm \sqrt{144 - 80}}{2 \cdot 1}$

Subtract $80$ from $144$ .

$y = \frac{- 12 \pm \sqrt{64}}{2 \cdot 1}$

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

$y = \frac{- 12 \pm \sqrt{8^{2}}}{2 \cdot 1}$

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

$y = \frac{- 12 \pm 8}{2 \cdot 1}$

Multiply $2$ by $1$ .

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

Simplify $\frac{- 12 \pm 8}{2}$ .

$y = - 6 \pm 4$

Step 4

The final answer is the combination of both solutions.

$y = - 2, - 10$