Solve Using the Quadratic Formula v^2-22v+21=0

$v^{2} - 22 v + 21 = 0$

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 $a = 1$ , $b = - 22$ , and $c = 21$ into the quadratic formula and solve for $v$ .

$\frac{22 \pm \sqrt{{(- 22)}^{2} - 4 \cdot (1 \cdot 21)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 22$ to the power of $2$ .

$v = \frac{22 \pm \sqrt{484 - 4 \cdot 1 \cdot 21}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 21$ .

Multiply $- 4$ by $1$ .

$v = \frac{22 \pm \sqrt{484 - 4 \cdot 21}}{2 \cdot 1}$

Multiply $- 4$ by $21$ .

$v = \frac{22 \pm \sqrt{484 - 84}}{2 \cdot 1}$

Subtract $84$ from $484$ .

$v = \frac{22 \pm \sqrt{400}}{2 \cdot 1}$

Rewrite $400$ as $20^{2}$ .

$v = \frac{22 \pm \sqrt{20^{2}}}{2 \cdot 1}$

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

$v = \frac{22 \pm 20}{2 \cdot 1}$

Multiply $2$ by $1$ .

$v = \frac{22 \pm 20}{2}$

Simplify $\frac{22 \pm 20}{2}$ .

$v = 11 \pm 10$

Step 4

The final answer is the combination of both solutions.

$v = 21, 1$