Precalculus Examples

$3 v^{2} + v + 3 = 0$

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 $a = 3$ , $b = 1$ , and $c = 3$ into the quadratic formula and solve for $v$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (3 \cdot 3)}}{2 \cdot 3}$

Step 3

Simplify.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

One to any power is one.

$v = \frac{- 1 \pm \sqrt{1 - 4 \cdot 3 \cdot 3}}{2 \cdot 3}$

Step 3.1.2

Multiply $- 4 \cdot 3 \cdot 3$ .

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Step 3.1.2.1

Multiply $- 4$ by $3$ .

$v = \frac{- 1 \pm \sqrt{1 - 12 \cdot 3}}{2 \cdot 3}$

Step 3.1.2.2

Multiply $- 12$ by $3$ .

$v = \frac{- 1 \pm \sqrt{1 - 36}}{2 \cdot 3}$

Step 3.1.3

Subtract $36$ from $1$ .

$v = \frac{- 1 \pm \sqrt{- 35}}{2 \cdot 3}$

Step 3.1.4

Rewrite $- 35$ as $- 1 (35)$ .

$v = \frac{- 1 \pm \sqrt{- 1 \cdot 35}}{2 \cdot 3}$

Step 3.1.5

Rewrite $\sqrt{- 1 (35)}$ as $\sqrt{- 1} \cdot \sqrt{35}$ .

$v = \frac{- 1 \pm \sqrt{- 1} \cdot \sqrt{35}}{2 \cdot 3}$

Step 3.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$v = \frac{- 1 \pm i \sqrt{35}}{2 \cdot 3}$

Step 3.2

Multiply $2$ by $3$ .

$v = \frac{- 1 \pm i \sqrt{35}}{6}$

Step 4

The final answer is the combination of both solutions.

$v = - \frac{1 - i \sqrt{35}}{6}, - \frac{1 + i \sqrt{35}}{6}$