Algebra Examples

x^{2} - x - 6 = 0

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

Step 1

Use the quadratic formula to find the solutions.

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

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

Step 2

Substitute the values

a = 1

$a = 1$ ,

b = - 1

$b = - 1$ , and

c = - 6

$c = - 6$ into the quadratic formula and solve for

x

$x$ .

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

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

Step 3

Simplify.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

Raise

- 1

$- 1$ to the power of

2

$2$ .

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

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

Step 3.1.2

Multiply

- 4 \cdot 1 \cdot - 6

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

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Step 3.1.2.1

Multiply

- 4

$- 4$ by

1

$1$ .

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

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

Step 3.1.2.2

Multiply

- 4

$- 4$ by

- 6

$- 6$ .

x = \frac{1 \pm \sqrt{1 + 24}}{2 \cdot 1}

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

x = \frac{1 \pm \sqrt{1 + 24}}{2 \cdot 1}

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

Step 3.1.3

Add

1

$1$ and

24

$24$ .

x = \frac{1 \pm \sqrt{25}}{2 \cdot 1}

$x = \frac{1 \pm \sqrt{25}}{2 \cdot 1}$

Step 3.1.4

Rewrite

25

$25$ as

5^{2}

$5^{2}$ .

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

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

Step 3.1.5

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

x = \frac{1 \pm 5}{2 \cdot 1}

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

x = \frac{1 \pm 5}{2 \cdot 1}

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

Step 3.2

Multiply

2

$2$ by

1

$1$ .

x = \frac{1 \pm 5}{2}

$x = \frac{1 \pm 5}{2}$

x = \frac{1 \pm 5}{2}

$x = \frac{1 \pm 5}{2}$

Step 4

The final answer is the combination of both solutions.

x = 3, - 2

$x = 3, - 2$