Solve Using the Quadratic Formula p^2+2p-63=0

$p^{2} + 2 p - 63 = 0$

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 $a = 1$ , $b = 2$ , and $c = - 63$ into the quadratic formula and solve for $p$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (1 \cdot - 63)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

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

$p = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 63}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 63$ .

Multiply $- 4$ by $1$ .

$p = \frac{- 2 \pm \sqrt{4 - 4 \cdot - 63}}{2 \cdot 1}$

Multiply $- 4$ by $- 63$ .

$p = \frac{- 2 \pm \sqrt{4 + 252}}{2 \cdot 1}$

Add $4$ and $252$ .

$p = \frac{- 2 \pm \sqrt{256}}{2 \cdot 1}$

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

$p = \frac{- 2 \pm \sqrt{16^{2}}}{2 \cdot 1}$

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

$p = \frac{- 2 \pm 16}{2 \cdot 1}$

Multiply $2$ by $1$ .

$p = \frac{- 2 \pm 16}{2}$

Simplify $\frac{- 2 \pm 16}{2}$ .

$p = - 1 \pm 8$

Step 4

The final answer is the combination of both solutions.

$p = 7, - 9$