Precalculus Examples

$3 x^{2} + 6 x - 5 = 0$

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 $a = 3$ , $b = 6$ , and $c = - 5$ into the quadratic formula and solve for $x$ .

$\frac{- 6 \pm \sqrt{6^{2} - 4 \cdot (3 \cdot - 5)}}{2 \cdot 3}$

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 $2$ .

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot 3 \cdot - 5}}{2 \cdot 3}$

Step 3.1.2

Multiply $- 4 \cdot 3 \cdot - 5$ .

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Step 3.1.2.1

Multiply $- 4$ by $3$ .

$x = \frac{- 6 \pm \sqrt{36 - 12 \cdot - 5}}{2 \cdot 3}$

Step 3.1.2.2

Multiply $- 12$ by $- 5$ .

$x = \frac{- 6 \pm \sqrt{36 + 60}}{2 \cdot 3}$

Step 3.1.3

Add $36$ and $60$ .

$x = \frac{- 6 \pm \sqrt{96}}{2 \cdot 3}$

Step 3.1.4

Rewrite $96$ as $4^{2} \cdot 6$ .

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Step 3.1.4.1

Factor $16$ out of $96$ .

$x = \frac{- 6 \pm \sqrt{16 (6)}}{2 \cdot 3}$

Step 3.1.4.2

Rewrite $16$ as $4^{2}$ .

$x = \frac{- 6 \pm \sqrt{4^{2} \cdot 6}}{2 \cdot 3}$

Step 3.1.5

Pull terms out from under the radical.

$x = \frac{- 6 \pm 4 \sqrt{6}}{2 \cdot 3}$

Step 3.2

Multiply $2$ by $3$ .

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

Step 3.3

Simplify $\frac{- 6 \pm 4 \sqrt{6}}{6}$ .

$x = \frac{- 3 \pm 2 \sqrt{6}}{3}$

Step 4

The final answer is the combination of both solutions.

$x = - \frac{3 - 2 \sqrt{6}}{3}, - \frac{3 + 2 \sqrt{6}}{3}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{3 - 2 \sqrt{6}}{3}, - \frac{3 + 2 \sqrt{6}}{3}$

Decimal Form:

$x = 0.63299316 \dots, - 2.63299316 \dots$