Basic Math Examples

$z^{2} + 4 z + 13 = 0$

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 $a = 1$ , $b = 4$ , and $c = 13$ into the quadratic formula and solve for $z$ .

$\frac{- 4 \pm \sqrt{4^{2} - 4 \cdot (1 \cdot 13)}}{2 \cdot 1}$

Step 3

Simplify.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

Raise $4$ to the power of $2$ .

$z = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot 13}}{2 \cdot 1}$

Step 3.1.2

Multiply $- 4 \cdot 1 \cdot 13$ .

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Step 3.1.2.1

Multiply $- 4$ by $1$ .

$z = \frac{- 4 \pm \sqrt{16 - 4 \cdot 13}}{2 \cdot 1}$

Step 3.1.2.2

Multiply $- 4$ by $13$ .

$z = \frac{- 4 \pm \sqrt{16 - 52}}{2 \cdot 1}$

Step 3.1.3

Subtract $52$ from $16$ .

$z = \frac{- 4 \pm \sqrt{- 36}}{2 \cdot 1}$

Step 3.1.4

Rewrite $- 36$ as $- 1 (36)$ .

$z = \frac{- 4 \pm \sqrt{- 1 \cdot 36}}{2 \cdot 1}$

Step 3.1.5

Rewrite $\sqrt{- 1 (36)}$ as $\sqrt{- 1} \cdot \sqrt{36}$ .

$z = \frac{- 4 \pm \sqrt{- 1} \cdot \sqrt{36}}{2 \cdot 1}$

Step 3.1.6

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

$z = \frac{- 4 \pm i \cdot \sqrt{36}}{2 \cdot 1}$

Step 3.1.7

Rewrite $36$ as $6^{2}$ .

$z = \frac{- 4 \pm i \cdot \sqrt{6^{2}}}{2 \cdot 1}$

Step 3.1.8

Pull terms out from under the radical, assuming positive real numbers.

$z = \frac{- 4 \pm i \cdot 6}{2 \cdot 1}$

Step 3.1.9

Move $6$ to the left of $i$ .

$z = \frac{- 4 \pm 6 i}{2 \cdot 1}$

Step 3.2

Multiply $2$ by $1$ .

$z = \frac{- 4 \pm 6 i}{2}$

Step 3.3

Simplify $\frac{- 4 \pm 6 i}{2}$ .

$z = - 2 \pm 3 i$

Step 4

The final answer is the combination of both solutions.

$z = - 2 + 3 i, - 2 - 3 i$