Algebra Examples

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

Step 1

Use the quadratic formula to find the solutions.

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

Step 2

Substitute the values $a = 6$ , $b = 5$ , and $c = - 6$ into the quadratic formula and solve for $x$ .

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

Step 3

Simplify.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

Raise $5$ to the power of $2$ .

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

Step 3.1.2

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

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Step 3.1.2.1

Multiply $- 4$ by $6$ .

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

Step 3.1.2.2

Multiply $- 24$ by $- 6$ .

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

Step 3.1.3

Add $25$ and $144$ .

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

Step 3.1.4

Rewrite $169$ as $13^{2}$ .

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

Step 3.1.5

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

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

Step 3.2

Multiply $2$ by $6$ .

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

Step 4

The final answer is the combination of both solutions.

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