Solve Using the Quadratic Formula y^2+13y+40=0

$y^{2} + 13 y + 40 = 0$

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 $a = 1$ , $b = 13$ , and $c = 40$ into the quadratic formula and solve for $y$ .

$\frac{- 13 \pm \sqrt{13^{2} - 4 \cdot (1 \cdot 40)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

Raise $13$ to the power of $2$ .

$y = \frac{- 13 \pm \sqrt{169 - 4 \cdot 1 \cdot 40}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 40$ .

Multiply $- 4$ by $1$ .

$y = \frac{- 13 \pm \sqrt{169 - 4 \cdot 40}}{2 \cdot 1}$

Multiply $- 4$ by $40$ .

$y = \frac{- 13 \pm \sqrt{169 - 160}}{2 \cdot 1}$

Subtract $160$ from $169$ .

$y = \frac{- 13 \pm \sqrt{9}}{2 \cdot 1}$

Rewrite $9$ as $3^{2}$ .

$y = \frac{- 13 \pm \sqrt{3^{2}}}{2 \cdot 1}$

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

$y = \frac{- 13 \pm 3}{2 \cdot 1}$

Multiply $2$ by $1$ .

$y = \frac{- 13 \pm 3}{2}$

Step 4

The final answer is the combination of both solutions.

$y = - 5, - 8$