Precalculus Examples

$3 x^{2} + 2 x = 7$

Step 1

Subtract $7$ from both sides of the equation.

$3 x^{2} + 2 x - 7 = 0$

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 $a = 3$ , $b = 2$ , and $c = - 7$ into the quadratic formula and solve for $x$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (3 \cdot - 7)}}{2 \cdot 3}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

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

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 3 \cdot - 7}}{2 \cdot 3}$

Step 4.1.2

Multiply $- 4 \cdot 3 \cdot - 7$ .

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Step 4.1.2.1

Multiply $- 4$ by $3$ .

$x = \frac{- 2 \pm \sqrt{4 - 12 \cdot - 7}}{2 \cdot 3}$

Step 4.1.2.2

Multiply $- 12$ by $- 7$ .

$x = \frac{- 2 \pm \sqrt{4 + 84}}{2 \cdot 3}$

Step 4.1.3

Add $4$ and $84$ .

$x = \frac{- 2 \pm \sqrt{88}}{2 \cdot 3}$

Step 4.1.4

Rewrite $88$ as $2^{2} \cdot 22$ .

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Step 4.1.4.1

Factor $4$ out of $88$ .

$x = \frac{- 2 \pm \sqrt{4 (22)}}{2 \cdot 3}$

Step 4.1.4.2

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

$x = \frac{- 2 \pm \sqrt{2^{2} \cdot 22}}{2 \cdot 3}$

Step 4.1.5

Pull terms out from under the radical.

$x = \frac{- 2 \pm 2 \sqrt{22}}{2 \cdot 3}$

Step 4.2

Multiply $2$ by $3$ .

$x = \frac{- 2 \pm 2 \sqrt{22}}{6}$

Step 4.3

Simplify $\frac{- 2 \pm 2 \sqrt{22}}{6}$ .

$x = \frac{- 1 \pm \sqrt{22}}{3}$

Step 5

The final answer is the combination of both solutions.

$x = - \frac{1 - \sqrt{22}}{3}, - \frac{1 + \sqrt{22}}{3}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{1 - \sqrt{22}}{3}, - \frac{1 + \sqrt{22}}{3}$

Decimal Form:

$x = 1.23013858 \dots, - 1.89680525 \dots$