Algebra Examples

6 x^{2} + 5 x - 6 = 0

$6 x^{2} + 5 x - 6 = 0$

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 = 6

$a = 6$ ,

b = 5

$b = 5$ , and

c = - 6

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

x

$x$ .

\frac{- 5 \pm \sqrt{5^{2} - 4 \cdot (6 \cdot - 6)}}{2 \cdot 6}

$\frac{- 5 \pm \sqrt{5^{2} - 4 \cdot (6 \cdot - 6)}}{2 \cdot 6}$

Step 3

Simplify.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

Raise

5

$5$ to the power of

2

$2$ .

x = \frac{- 5 \pm \sqrt{25 - 4 \cdot 6 \cdot - 6}}{2 \cdot 6}

$x = \frac{- 5 \pm \sqrt{25 - 4 \cdot 6 \cdot - 6}}{2 \cdot 6}$

Step 3.1.2

Multiply

- 4 \cdot 6 \cdot - 6

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

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Step 3.1.2.1

Multiply

- 4

$- 4$ by

6

$6$ .

x = \frac{- 5 \pm \sqrt{25 - 24 \cdot - 6}}{2 \cdot 6}

$x = \frac{- 5 \pm \sqrt{25 - 24 \cdot - 6}}{2 \cdot 6}$

Step 3.1.2.2

Multiply

- 24

$- 24$ by

- 6

$- 6$ .

x = \frac{- 5 \pm \sqrt{25 + 144}}{2 \cdot 6}

$x = \frac{- 5 \pm \sqrt{25 + 144}}{2 \cdot 6}$

x = \frac{- 5 \pm \sqrt{25 + 144}}{2 \cdot 6}

$x = \frac{- 5 \pm \sqrt{25 + 144}}{2 \cdot 6}$

Step 3.1.3

Add

25

$25$ and

144

$144$ .

x = \frac{- 5 \pm \sqrt{169}}{2 \cdot 6}

$x = \frac{- 5 \pm \sqrt{169}}{2 \cdot 6}$

Step 3.1.4

Rewrite

169

$169$ as

13^{2}

$13^{2}$ .

x = \frac{- 5 \pm \sqrt{13^{2}}}{2 \cdot 6}

$x = \frac{- 5 \pm \sqrt{13^{2}}}{2 \cdot 6}$

Step 3.1.5

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

x = \frac{- 5 \pm 13}{2 \cdot 6}

$x = \frac{- 5 \pm 13}{2 \cdot 6}$

x = \frac{- 5 \pm 13}{2 \cdot 6}

$x = \frac{- 5 \pm 13}{2 \cdot 6}$

Step 3.2

Multiply

2

$2$ by

6

$6$ .

x = \frac{- 5 \pm 13}{12}

$x = \frac{- 5 \pm 13}{12}$

x = \frac{- 5 \pm 13}{12}

$x = \frac{- 5 \pm 13}{12}$

Step 4

The final answer is the combination of both solutions.

x = \frac{2}{3}, - \frac{3}{2}

$x = \frac{2}{3}, - \frac{3}{2}$