Find All Complex Solutions 2x(x-3)=0

$2 x (x - 3) = 0$

Step 1

Simplify the left side.

Simplify $2 x (x - 3)$ .

Apply the distributive property.

$2 x \cdot x + 2 x \cdot - 3 = 0$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$2 (x \cdot x) + 2 x \cdot - 3 = 0$

Multiply $x$ by $x$ .

$2 x^{2} + 2 x \cdot - 3 = 0$

Multiply $- 3$ by $2$ .

$2 x^{2} - 6 x = 0$

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 $a = 2$ , $b = - 6$ , and $c = 0$ into the quadratic formula and solve for $x$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (2 \cdot 0)}}{2 \cdot 2}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 6$ to the power of $2$ .

$x = \frac{6 \pm \sqrt{36 - 4 \cdot 2 \cdot 0}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot 0$ .

Multiply $- 4$ by $2$ .

$x = \frac{6 \pm \sqrt{36 - 8 \cdot 0}}{2 \cdot 2}$

Multiply $- 8$ by $0$ .

$x = \frac{6 \pm \sqrt{36 + 0}}{2 \cdot 2}$

Add $36$ and $0$ .

$x = \frac{6 \pm \sqrt{36}}{2 \cdot 2}$

Rewrite $36$ as $6^{2}$ .

$x = \frac{6 \pm \sqrt{6^{2}}}{2 \cdot 2}$

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

$x = \frac{6 \pm 6}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{6 \pm 6}{4}$

Simplify $\frac{6 \pm 6}{4}$ .

$x = \frac{3 \pm 3}{2}$

Step 5

The final answer is the combination of both solutions.

$x = 3, 0$