Algebra Examples

5 x^{2} - 2 x - 7 = 0

$5 x^{2} - 2 x - 7 = 0$

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 = 5

$a = 5$ ,

b = - 2

$b = - 2$ , and

c = - 7

$c = - 7$ into the quadratic formula and solve for

x

$x$ .

\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (5 \cdot - 7)}}{2 \cdot 5}

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (5 \cdot - 7)}}{2 \cdot 5}$

Step 3

Simplify.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

Raise

- 2

$- 2$ to the power of

2

$2$ .

x = \frac{2 \pm \sqrt{4 - 4 \cdot 5 \cdot - 7}}{2 \cdot 5}

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 5 \cdot - 7}}{2 \cdot 5}$

Step 3.1.2

Multiply

- 4 \cdot 5 \cdot - 7

$- 4 \cdot 5 \cdot - 7$ .

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Step 3.1.2.1

Multiply

- 4

$- 4$ by

5

$5$ .

x = \frac{2 \pm \sqrt{4 - 20 \cdot - 7}}{2 \cdot 5}

$x = \frac{2 \pm \sqrt{4 - 20 \cdot - 7}}{2 \cdot 5}$

Step 3.1.2.2

Multiply

- 20

$- 20$ by

- 7

$- 7$ .

x = \frac{2 \pm \sqrt{4 + 140}}{2 \cdot 5}

$x = \frac{2 \pm \sqrt{4 + 140}}{2 \cdot 5}$

x = \frac{2 \pm \sqrt{4 + 140}}{2 \cdot 5}

$x = \frac{2 \pm \sqrt{4 + 140}}{2 \cdot 5}$

Step 3.1.3

Add

4

$4$ and

140

$140$ .

x = \frac{2 \pm \sqrt{144}}{2 \cdot 5}

$x = \frac{2 \pm \sqrt{144}}{2 \cdot 5}$

Step 3.1.4

Rewrite

144

$144$ as

12^{2}

$12^{2}$ .

x = \frac{2 \pm \sqrt{12^{2}}}{2 \cdot 5}

$x = \frac{2 \pm \sqrt{12^{2}}}{2 \cdot 5}$

Step 3.1.5

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

x = \frac{2 \pm 12}{2 \cdot 5}

$x = \frac{2 \pm 12}{2 \cdot 5}$

x = \frac{2 \pm 12}{2 \cdot 5}

$x = \frac{2 \pm 12}{2 \cdot 5}$

Step 3.2

Multiply

2

$2$ by

5

$5$ .

x = \frac{2 \pm 12}{10}

$x = \frac{2 \pm 12}{10}$

Step 3.3

Simplify

\frac{2 \pm 12}{10}

$\frac{2 \pm 12}{10}$ .

x = \frac{1 \pm 6}{5}

$x = \frac{1 \pm 6}{5}$

x = \frac{1 \pm 6}{5}

$x = \frac{1 \pm 6}{5}$

Step 4

The final answer is the combination of both solutions.

x = \frac{7}{5}, - 1

$x = \frac{7}{5}, - 1$