Finite Math Examples

$- x^{2} - 2 x - 6$

Step 1

Set $- x^{2} - 2 x - 6$ equal to $0$ .

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

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3

Substitute the values $a = - 1$ , $b = - 2$ , and $c = - 6$ into the quadratic formula and solve for $x$ .

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

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

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

$x = \frac{2 \pm \sqrt{4 - 4 \cdot - 1 \cdot - 6}}{2 \cdot - 1}$

Step 4.1.2

Multiply $- 4 \cdot - 1 \cdot - 6$ .

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Step 4.1.2.1

Multiply $- 4$ by $- 1$ .

$x = \frac{2 \pm \sqrt{4 + 4 \cdot - 6}}{2 \cdot - 1}$

Step 4.1.2.2

Multiply $4$ by $- 6$ .

$x = \frac{2 \pm \sqrt{4 - 24}}{2 \cdot - 1}$

Step 4.1.3

Subtract $24$ from $4$ .

$x = \frac{2 \pm \sqrt{- 20}}{2 \cdot - 1}$

Step 4.1.4

Rewrite $- 20$ as $- 1 (20)$ .

$x = \frac{2 \pm \sqrt{- 1 \cdot 20}}{2 \cdot - 1}$

Step 4.1.5

Rewrite $\sqrt{- 1 (20)}$ as $\sqrt{- 1} \cdot \sqrt{20}$ .

$x = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{20}}{2 \cdot - 1}$

Step 4.1.6

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

$x = \frac{2 \pm i \cdot \sqrt{20}}{2 \cdot - 1}$

Step 4.1.7

Rewrite $20$ as $2^{2} \cdot 5$ .

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Step 4.1.7.1

Factor $4$ out of $20$ .

$x = \frac{2 \pm i \cdot \sqrt{4 (5)}}{2 \cdot - 1}$

Step 4.1.7.2

Rewrite $4$ as $2^{2}$ .

$x = \frac{2 \pm i \cdot \sqrt{2^{2} \cdot 5}}{2 \cdot - 1}$

Step 4.1.8

Pull terms out from under the radical.

$x = \frac{2 \pm i \cdot (2 \sqrt{5})}{2 \cdot - 1}$

Step 4.1.9

Move $2$ to the left of $i$ .

$x = \frac{2 \pm 2 i \sqrt{5}}{2 \cdot - 1}$

Step 4.2

Multiply $2$ by $- 1$ .

$x = \frac{2 \pm 2 i \sqrt{5}}{- 2}$

Step 4.3

Simplify $\frac{2 \pm 2 i \sqrt{5}}{- 2}$ .

$x = \frac{1 \pm i \sqrt{5}}{- 1}$

Step 4.4

Move the negative one from the denominator of $\frac{1 \pm i \sqrt{5}}{- 1}$ .

$x = - 1 \cdot (1 \pm i \sqrt{5})$

Step 4.5

Rewrite $- 1 \cdot (1 \pm i \sqrt{5})$ as $- (1 \pm i \sqrt{5})$ .

$x = - (1 \pm i \sqrt{5})$