Solve Using the Quadratic Formula -p^2-11p=0

$- p^{2} - 11 p = 0$

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 $a = - 1$ , $b = - 11$ , and $c = 0$ into the quadratic formula and solve for $p$ .

$\frac{11 \pm \sqrt{{(- 11)}^{2} - 4 \cdot (- 1 \cdot 0)}}{2 \cdot - 1}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 11$ to the power of $2$ .

$p = \frac{11 \pm \sqrt{121 - 4 \cdot - 1 \cdot 0}}{2 \cdot - 1}$

Multiply $- 4 \cdot - 1 \cdot 0$ .

Multiply $- 4$ by $- 1$ .

$p = \frac{11 \pm \sqrt{121 + 4 \cdot 0}}{2 \cdot - 1}$

Multiply $4$ by $0$ .

$p = \frac{11 \pm \sqrt{121 + 0}}{2 \cdot - 1}$

Add $121$ and $0$ .

$p = \frac{11 \pm \sqrt{121}}{2 \cdot - 1}$

Rewrite $121$ as $11^{2}$ .

$p = \frac{11 \pm \sqrt{11^{2}}}{2 \cdot - 1}$

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

$p = \frac{11 \pm 11}{2 \cdot - 1}$

Multiply $2$ by $- 1$ .

$p = \frac{11 \pm 11}{- 2}$

Move the negative in front of the fraction.

$p = - \frac{11 \pm 11}{2}$

Step 4

The final answer is the combination of both solutions.

$p = - 11, 0$