Solve Using the Quadratic Formula 9x^2+15x=0

$9 x^{2} + 15 x = 0$

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 $a = 9$ , $b = 15$ , and $c = 0$ into the quadratic formula and solve for $x$ .

$\frac{- 15 \pm \sqrt{15^{2} - 4 \cdot (9 \cdot 0)}}{2 \cdot 9}$

Step 3

Simplify.

Simplify the numerator.

Raise $15$ to the power of $2$ .

$x = \frac{- 15 \pm \sqrt{225 - 4 \cdot 9 \cdot 0}}{2 \cdot 9}$

Multiply $- 4 \cdot 9 \cdot 0$ .

Multiply $- 4$ by $9$ .

$x = \frac{- 15 \pm \sqrt{225 - 36 \cdot 0}}{2 \cdot 9}$

Multiply $- 36$ by $0$ .

$x = \frac{- 15 \pm \sqrt{225 + 0}}{2 \cdot 9}$

Add $225$ and $0$ .

$x = \frac{- 15 \pm \sqrt{225}}{2 \cdot 9}$

Rewrite $225$ as $15^{2}$ .

$x = \frac{- 15 \pm \sqrt{15^{2}}}{2 \cdot 9}$

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

$x = \frac{- 15 \pm 15}{2 \cdot 9}$

Multiply $2$ by $9$ .

$x = \frac{- 15 \pm 15}{18}$

Simplify $\frac{- 15 \pm 15}{18}$ .

$x = \frac{- 5 \pm 5}{6}$

Step 4

The final answer is the combination of both solutions.

$x = 0, - \frac{5}{3}$