Solve Using the Quadratic Formula v^2-9v+18=0

$v^{2} - 9 v + 18 = 0$

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 $a = 1$ , $b = - 9$ , and $c = 18$ into the quadratic formula and solve for $v$ .

$\frac{9 \pm \sqrt{{(- 9)}^{2} - 4 \cdot (1 \cdot 18)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 9$ to the power of $2$ .

$v = \frac{9 \pm \sqrt{81 - 4 \cdot 1 \cdot 18}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 18$ .

Multiply $- 4$ by $1$ .

$v = \frac{9 \pm \sqrt{81 - 4 \cdot 18}}{2 \cdot 1}$

Multiply $- 4$ by $18$ .

$v = \frac{9 \pm \sqrt{81 - 72}}{2 \cdot 1}$

Subtract $72$ from $81$ .

$v = \frac{9 \pm \sqrt{9}}{2 \cdot 1}$

Rewrite $9$ as $3^{2}$ .

$v = \frac{9 \pm \sqrt{3^{2}}}{2 \cdot 1}$

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

$v = \frac{9 \pm 3}{2 \cdot 1}$

Multiply $2$ by $1$ .

$v = \frac{9 \pm 3}{2}$

Step 4

The final answer is the combination of both solutions.

$v = 6, 3$