Basic Math Examples

$a^{2} - 2 a + 4 = 0$

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 $a = 1$ , $b = - 2$ , and $c = 4$ into the quadratic formula and solve for $a$ .

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

Step 3

Simplify.

Tap for more steps...

Step 3.1

Simplify the numerator.

Tap for more steps...

Step 3.1.1

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

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

Step 3.1.2

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

Tap for more steps...

Step 3.1.2.1

Multiply $- 4$ by $1$ .

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

Step 3.1.2.2

Multiply $- 4$ by $4$ .

$a = \frac{2 \pm \sqrt{4 - 16}}{2 \cdot 1}$

Step 3.1.3

Subtract $16$ from $4$ .

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

Step 3.1.4

Rewrite $- 12$ as $- 1 (12)$ .

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

Step 3.1.5

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

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

Step 3.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$a = \frac{2 \pm i \cdot \sqrt{12}}{2 \cdot 1}$

Step 3.1.7

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

Tap for more steps...

Step 3.1.7.1

Factor $4$ out of $12$ .

$a = \frac{2 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 1}$

Step 3.1.7.2

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

$a = \frac{2 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

Step 3.1.8

Pull terms out from under the radical.

$a = \frac{2 \pm i \cdot (2 \sqrt{3})}{2 \cdot 1}$

Step 3.1.9

Move $2$ to the left of $i$ .

$a = \frac{2 \pm 2 i \sqrt{3}}{2 \cdot 1}$

Step 3.2

Multiply $2$ by $1$ .

$a = \frac{2 \pm 2 i \sqrt{3}}{2}$

Step 3.3

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

$a = 1 \pm i \sqrt{3}$

Step 4

The final answer is the combination of both solutions.

$a = 1 + i \sqrt{3}, 1 - i \sqrt{3}$