Solve Using the Quadratic Formula m^2+9m=0

$m^{2} + 9 m = 0$

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 $a = 1$ , $b = 9$ , and $c = 0$ into the quadratic formula and solve for $m$ .

$\frac{- 9 \pm \sqrt{9^{2} - 4 \cdot (1 \cdot 0)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

Raise $9$ to the power of $2$ .

$m = \frac{- 9 \pm \sqrt{81 - 4 \cdot 1 \cdot 0}}{2 \cdot 1}$

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

Multiply $- 4$ by $1$ .

$m = \frac{- 9 \pm \sqrt{81 - 4 \cdot 0}}{2 \cdot 1}$

Multiply $- 4$ by $0$ .

$m = \frac{- 9 \pm \sqrt{81 + 0}}{2 \cdot 1}$

Add $81$ and $0$ .

$m = \frac{- 9 \pm \sqrt{81}}{2 \cdot 1}$

Rewrite $81$ as $9^{2}$ .

$m = \frac{- 9 \pm \sqrt{9^{2}}}{2 \cdot 1}$

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

$m = \frac{- 9 \pm 9}{2 \cdot 1}$

Multiply $2$ by $1$ .

$m = \frac{- 9 \pm 9}{2}$

Step 4

The final answer is the combination of both solutions.

$m = 0, - 9$