Solve Using the Quadratic Formula a^2+14a+40=0

$a^{2} + 14 a + 40 = 0$

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 $a = 1$ , $b = 14$ , and $c = 40$ into the quadratic formula and solve for $a$ .

$\frac{- 14 \pm \sqrt{14^{2} - 4 \cdot (1 \cdot 40)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

Raise $14$ to the power of $2$ .

$a = \frac{- 14 \pm \sqrt{196 - 4 \cdot 1 \cdot 40}}{2 \cdot 1}$

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

Multiply $- 4$ by $1$ .

$a = \frac{- 14 \pm \sqrt{196 - 4 \cdot 40}}{2 \cdot 1}$

Multiply $- 4$ by $40$ .

$a = \frac{- 14 \pm \sqrt{196 - 160}}{2 \cdot 1}$

Subtract $160$ from $196$ .

$a = \frac{- 14 \pm \sqrt{36}}{2 \cdot 1}$

Rewrite $36$ as $6^{2}$ .

$a = \frac{- 14 \pm \sqrt{6^{2}}}{2 \cdot 1}$

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

$a = \frac{- 14 \pm 6}{2 \cdot 1}$

Multiply $2$ by $1$ .

$a = \frac{- 14 \pm 6}{2}$

Simplify $\frac{- 14 \pm 6}{2}$ .

$a = - 7 \pm 3$

Step 4

The final answer is the combination of both solutions.

$a = - 4, - 10$