Find All Complex Solutions (21x^2+14x)=0

$(21 x^{2} + 14 x) = 0$

Step 1

Simplify the left side.

Remove parentheses.

$21 x^{2} + 14 x = 0$

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3

Substitute the values $a = 21$ , $b = 14$ , and $c = 0$ into the quadratic formula and solve for $x$ .

$\frac{- 14 \pm \sqrt{14^{2} - 4 \cdot (21 \cdot 0)}}{2 \cdot 21}$

Step 4

Simplify.

Simplify the numerator.

Raise $14$ to the power of $2$ .

$x = \frac{- 14 \pm \sqrt{196 - 4 \cdot 21 \cdot 0}}{2 \cdot 21}$

Multiply $- 4 \cdot 21 \cdot 0$ .

Multiply $- 4$ by $21$ .

$x = \frac{- 14 \pm \sqrt{196 - 84 \cdot 0}}{2 \cdot 21}$

Multiply $- 84$ by $0$ .

$x = \frac{- 14 \pm \sqrt{196 + 0}}{2 \cdot 21}$

Add $196$ and $0$ .

$x = \frac{- 14 \pm \sqrt{196}}{2 \cdot 21}$

Rewrite $196$ as $14^{2}$ .

$x = \frac{- 14 \pm \sqrt{14^{2}}}{2 \cdot 21}$

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

$x = \frac{- 14 \pm 14}{2 \cdot 21}$

Multiply $2$ by $21$ .

$x = \frac{- 14 \pm 14}{42}$

Simplify $\frac{- 14 \pm 14}{42}$ .

$x = \frac{- 1 \pm 1}{3}$

Step 5

The final answer is the combination of both solutions.

$x = 0, - \frac{2}{3}$