Solve Using the Quadratic Formula v^2-6v=0

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

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 $a = 1$ , $b = - 6$ , and $c = 0$ into the quadratic formula and solve for $v$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (1 \cdot 0)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

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

$v = \frac{6 \pm \sqrt{36 - 4 \cdot 1 \cdot 0}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 0$ .

Multiply $- 4$ by $1$ .

$v = \frac{6 \pm \sqrt{36 - 4 \cdot 0}}{2 \cdot 1}$

Multiply $- 4$ by $0$ .

$v = \frac{6 \pm \sqrt{36 + 0}}{2 \cdot 1}$

Add $36$ and $0$ .

$v = \frac{6 \pm \sqrt{36}}{2 \cdot 1}$

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

$v = \frac{6 \pm \sqrt{6^{2}}}{2 \cdot 1}$

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

$v = \frac{6 \pm 6}{2 \cdot 1}$

Multiply $2$ by $1$ .

$v = \frac{6 \pm 6}{2}$

Simplify $\frac{6 \pm 6}{2}$ .

$v = 3 \pm 3$

Step 4

The final answer is the combination of both solutions.

$v = 6, 0$