Pre-Algebra Examples

$x^{2} + 6 x - 1 = 0$

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

$a = 1$ ,

$b = 6$ , and

$c = - 1$ into the quadratic formula and solve for

$x$ .

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

Step 3

Simplify.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

Raise

$6$ to the power of

$2$ .

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot 1 \cdot - 1}}{2 \cdot 1}$

Step 3.1.2

Multiply

$- 4 \cdot 1 \cdot - 1$ .

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Step 3.1.2.1

Multiply

$- 4$ by

$1$ .

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot - 1}}{2 \cdot 1}$

Step 3.1.2.2

Multiply

$- 4$ by

$- 1$ .

$x = \frac{- 6 \pm \sqrt{36 + 4}}{2 \cdot 1}$

Step 3.1.3

Add

$36$ and

$4$ .

$x = \frac{- 6 \pm \sqrt{40}}{2 \cdot 1}$

Step 3.1.4

Rewrite

$40$ as

$2^{2} \cdot 10$ .

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Step 3.1.4.1

Factor

$4$ out of

$40$ .

$x = \frac{- 6 \pm \sqrt{4 (10)}}{2 \cdot 1}$

Step 3.1.4.2

Rewrite

$4$ as

$2^{2}$ .

$x = \frac{- 6 \pm \sqrt{2^{2} \cdot 10}}{2 \cdot 1}$

Step 3.1.5

Pull terms out from under the radical.

$x = \frac{- 6 \pm 2 \sqrt{10}}{2 \cdot 1}$

Step 3.2

Multiply

$2$ by

$1$ .

$x = \frac{- 6 \pm 2 \sqrt{10}}{2}$

Step 3.3

Simplify

$\frac{- 6 \pm 2 \sqrt{10}}{2}$ .

$x = - 3 \pm \sqrt{10}$

Step 4

Simplify the expression to solve for the

$+$ portion of the

$\pm$ .

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Raise

$6$ to the power of

$2$ .

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot 1 \cdot - 1}}{2 \cdot 1}$

Step 4.1.2

Multiply

$- 4 \cdot 1 \cdot - 1$ .

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Step 4.1.2.1

Multiply

$- 4$ by

$1$ .

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot - 1}}{2 \cdot 1}$

Step 4.1.2.2

Multiply

$- 4$ by

$- 1$ .

$x = \frac{- 6 \pm \sqrt{36 + 4}}{2 \cdot 1}$

Step 4.1.3

Add

$36$ and

$4$ .

$x = \frac{- 6 \pm \sqrt{40}}{2 \cdot 1}$

Step 4.1.4

Rewrite

$40$ as

$2^{2} \cdot 10$ .

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Step 4.1.4.1

Factor

$4$ out of

$40$ .

$x = \frac{- 6 \pm \sqrt{4 (10)}}{2 \cdot 1}$

Step 4.1.4.2

Rewrite

$4$ as

$2^{2}$ .

$x = \frac{- 6 \pm \sqrt{2^{2} \cdot 10}}{2 \cdot 1}$

Step 4.1.5

Pull terms out from under the radical.

$x = \frac{- 6 \pm 2 \sqrt{10}}{2 \cdot 1}$

Step 4.2

Multiply

$2$ by

$1$ .

$x = \frac{- 6 \pm 2 \sqrt{10}}{2}$

Step 4.3

Simplify

$\frac{- 6 \pm 2 \sqrt{10}}{2}$ .

$x = - 3 \pm \sqrt{10}$

Step 4.4

Change the

$\pm$ to

$+$ .

$x = - 3 + \sqrt{10}$

Step 5

Simplify the expression to solve for the

$-$ portion of the

$\pm$ .

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Step 5.1

Simplify the numerator.

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Step 5.1.1

Raise

$6$ to the power of

$2$ .

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot 1 \cdot - 1}}{2 \cdot 1}$

Step 5.1.2

Multiply

$- 4 \cdot 1 \cdot - 1$ .

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Step 5.1.2.1

Multiply

$- 4$ by

$1$ .

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot - 1}}{2 \cdot 1}$

Step 5.1.2.2

Multiply

$- 4$ by

$- 1$ .

$x = \frac{- 6 \pm \sqrt{36 + 4}}{2 \cdot 1}$

Step 5.1.3

Add

$36$ and

$4$ .

$x = \frac{- 6 \pm \sqrt{40}}{2 \cdot 1}$

Step 5.1.4

Rewrite

$40$ as

$2^{2} \cdot 10$ .

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Step 5.1.4.1

Factor

$4$ out of

$40$ .

$x = \frac{- 6 \pm \sqrt{4 (10)}}{2 \cdot 1}$

Step 5.1.4.2

Rewrite

$4$ as

$2^{2}$ .

$x = \frac{- 6 \pm \sqrt{2^{2} \cdot 10}}{2 \cdot 1}$

Step 5.1.5

Pull terms out from under the radical.

$x = \frac{- 6 \pm 2 \sqrt{10}}{2 \cdot 1}$

Step 5.2

Multiply

$2$ by

$1$ .

$x = \frac{- 6 \pm 2 \sqrt{10}}{2}$

Step 5.3

Simplify

$\frac{- 6 \pm 2 \sqrt{10}}{2}$ .

$x = - 3 \pm \sqrt{10}$

Step 5.4

Change the

$\pm$ to

$-$ .

$x = - 3 - \sqrt{10}$

Step 6

The final answer is the combination of both solutions.

$x = - 3 + \sqrt{10}, - 3 - \sqrt{10}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$x = - 3 + \sqrt{10}, - 3 - \sqrt{10}$

Decimal Form:

$x = 0.16227766 \dots, - 6.16227766 \dots$