Precalculus Examples

$f (x) = x^{2} + 2 x - 2$

Step 1

Set $x^{2} + 2 x - 2$ equal to $0$ .

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

Step 2

Solve for $x$ .

Tap for more steps...

Step 2.1

Use the quadratic formula to find the solutions.

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

Step 2.2

Substitute the values $a = 1$ , $b = 2$ , and $c = - 2$ into the quadratic formula and solve for $x$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (1 \cdot - 2)}}{2 \cdot 1}$

Step 2.3

Simplify.

Tap for more steps...

Step 2.3.1

Simplify the numerator.

Tap for more steps...

Step 2.3.1.1

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

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 2}}{2 \cdot 1}$

Step 2.3.1.2

Multiply $- 4 \cdot 1 \cdot - 2$ .

Tap for more steps...

Step 2.3.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot - 2}}{2 \cdot 1}$

Step 2.3.1.2.2

Multiply $- 4$ by $- 2$ .

$x = \frac{- 2 \pm \sqrt{4 + 8}}{2 \cdot 1}$

Step 2.3.1.3

Add $4$ and $8$ .

$x = \frac{- 2 \pm \sqrt{12}}{2 \cdot 1}$

Step 2.3.1.4

Rewrite $12$ as $2^{2} \cdot 3$ .

Tap for more steps...

Step 2.3.1.4.1

Factor $4$ out of $12$ .

$x = \frac{- 2 \pm \sqrt{4 (3)}}{2 \cdot 1}$

Step 2.3.1.4.2

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

$x = \frac{- 2 \pm \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

Step 2.3.1.5

Pull terms out from under the radical.

$x = \frac{- 2 \pm 2 \sqrt{3}}{2 \cdot 1}$

Step 2.3.2

Multiply $2$ by $1$ .

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

Step 2.3.3

Simplify $\frac{- 2 \pm 2 \sqrt{3}}{2}$ .

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

Step 2.4

The final answer is the combination of both solutions.

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

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

Step 3

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$x = 0.73205080 \dots, - 2.73205080 \dots$

Step 4