Find All Complex Solutions 7x^2-42x=0

$7 x^{2} - 42 x = 0$

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 $a = 7$ , $b = - 42$ , and $c = 0$ into the quadratic formula and solve for $x$ .

$\frac{42 \pm \sqrt{{(- 42)}^{2} - 4 \cdot (7 \cdot 0)}}{2 \cdot 7}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 42$ to the power of $2$ .

$x = \frac{42 \pm \sqrt{1764 - 4 \cdot 7 \cdot 0}}{2 \cdot 7}$

Multiply $- 4 \cdot 7 \cdot 0$ .

Multiply $- 4$ by $7$ .

$x = \frac{42 \pm \sqrt{1764 - 28 \cdot 0}}{2 \cdot 7}$

Multiply $- 28$ by $0$ .

$x = \frac{42 \pm \sqrt{1764 + 0}}{2 \cdot 7}$

Add $1764$ and $0$ .

$x = \frac{42 \pm \sqrt{1764}}{2 \cdot 7}$

Rewrite $1764$ as $42^{2}$ .

$x = \frac{42 \pm \sqrt{42^{2}}}{2 \cdot 7}$

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

$x = \frac{42 \pm 42}{2 \cdot 7}$

Multiply $2$ by $7$ .

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

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

$x = 3 \pm 3$

Step 4

The final answer is the combination of both solutions.

$x = 6, 0$