Precalculus Examples

$5 x^{2} = - 12 x - 2$

Step 1

Move all the expressions to the left side of the equation.

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Step 1.1

Add $12 x$ to both sides of the equation.

$5 x^{2} + 12 x = - 2$

Step 1.2

Add $2$ to both sides of the equation.

$5 x^{2} + 12 x + 2 = 0$

Step 2

Use the quadratic formula to find the solutions.

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

Step 3

Substitute the values $a = 5$ , $b = 12$ , and $c = 2$ into the quadratic formula and solve for $x$ .

$\frac{- 12 \pm \sqrt{12^{2} - 4 \cdot (5 \cdot 2)}}{2 \cdot 5}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Raise $12$ to the power of $2$ .

$x = \frac{- 12 \pm \sqrt{144 - 4 \cdot 5 \cdot 2}}{2 \cdot 5}$

Step 4.1.2

Multiply $- 4 \cdot 5 \cdot 2$ .

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Step 4.1.2.1

Multiply $- 4$ by $5$ .

$x = \frac{- 12 \pm \sqrt{144 - 20 \cdot 2}}{2 \cdot 5}$

Step 4.1.2.2

Multiply $- 20$ by $2$ .

$x = \frac{- 12 \pm \sqrt{144 - 40}}{2 \cdot 5}$

Step 4.1.3

Subtract $40$ from $144$ .

$x = \frac{- 12 \pm \sqrt{104}}{2 \cdot 5}$

Step 4.1.4

Rewrite $104$ as $2^{2} \cdot 26$ .

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Step 4.1.4.1

Factor $4$ out of $104$ .

$x = \frac{- 12 \pm \sqrt{4 (26)}}{2 \cdot 5}$

Step 4.1.4.2

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

$x = \frac{- 12 \pm \sqrt{2^{2} \cdot 26}}{2 \cdot 5}$

Step 4.1.5

Pull terms out from under the radical.

$x = \frac{- 12 \pm 2 \sqrt{26}}{2 \cdot 5}$

Step 4.2

Multiply $2$ by $5$ .

$x = \frac{- 12 \pm 2 \sqrt{26}}{10}$

Step 4.3

Simplify $\frac{- 12 \pm 2 \sqrt{26}}{10}$ .

$x = \frac{- 6 \pm \sqrt{26}}{5}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = \frac{- 6 \pm \sqrt{26}}{5}$

Decimal Form:

$x = - 0.18019609 \dots, - 2.21980390 \dots$