Solve Using the Quadratic Formula 15x^2+13x=0

$15 x^{2} + 13 x = 0$

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 $a = 15$ , $b = 13$ , and $c = 0$ into the quadratic formula and solve for $x$ .

$\frac{- 13 \pm \sqrt{13^{2} - 4 \cdot (15 \cdot 0)}}{2 \cdot 15}$

Step 3

Simplify.

Simplify the numerator.

Raise $13$ to the power of $2$ .

$x = \frac{- 13 \pm \sqrt{169 - 4 \cdot 15 \cdot 0}}{2 \cdot 15}$

Multiply $- 4 \cdot 15 \cdot 0$ .

Multiply $- 4$ by $15$ .

$x = \frac{- 13 \pm \sqrt{169 - 60 \cdot 0}}{2 \cdot 15}$

Multiply $- 60$ by $0$ .

$x = \frac{- 13 \pm \sqrt{169 + 0}}{2 \cdot 15}$

Add $169$ and $0$ .

$x = \frac{- 13 \pm \sqrt{169}}{2 \cdot 15}$

Rewrite $169$ as $13^{2}$ .

$x = \frac{- 13 \pm \sqrt{13^{2}}}{2 \cdot 15}$

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

$x = \frac{- 13 \pm 13}{2 \cdot 15}$

Multiply $2$ by $15$ .

$x = \frac{- 13 \pm 13}{30}$

Step 4

The final answer is the combination of both solutions.

$x = 0, - \frac{13}{15}$