Solve Using the Quadratic Formula 5v^2+29v+20=0

$5 v^{2} + 29 v + 20 = 0$

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 $a = 5$ , $b = 29$ , and $c = 20$ into the quadratic formula and solve for $v$ .

$\frac{- 29 \pm \sqrt{29^{2} - 4 \cdot (5 \cdot 20)}}{2 \cdot 5}$

Step 3

Simplify.

Simplify the numerator.

Raise $29$ to the power of $2$ .

$v = \frac{- 29 \pm \sqrt{841 - 4 \cdot 5 \cdot 20}}{2 \cdot 5}$

Multiply $- 4 \cdot 5 \cdot 20$ .

Multiply $- 4$ by $5$ .

$v = \frac{- 29 \pm \sqrt{841 - 20 \cdot 20}}{2 \cdot 5}$

Multiply $- 20$ by $20$ .

$v = \frac{- 29 \pm \sqrt{841 - 400}}{2 \cdot 5}$

Subtract $400$ from $841$ .

$v = \frac{- 29 \pm \sqrt{441}}{2 \cdot 5}$

Rewrite $441$ as $21^{2}$ .

$v = \frac{- 29 \pm \sqrt{21^{2}}}{2 \cdot 5}$

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

$v = \frac{- 29 \pm 21}{2 \cdot 5}$

Multiply $2$ by $5$ .

$v = \frac{- 29 \pm 21}{10}$

Step 4

The final answer is the combination of both solutions.

$v = - \frac{4}{5}, - 5$