Algebra Examples

$14 - z^{2} = 17$

Step 1

Move all terms not containing $z$ to the right side of the equation.

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Step 1.1

Subtract $14$ from both sides of the equation.

$- z^{2} = 17 - 14$

Step 1.2

Subtract $14$ from $17$ .

$- z^{2} = 3$

Step 2

Divide each term in $- z^{2} = 3$ by $- 1$ and simplify.

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Step 2.1

Divide each term in $- z^{2} = 3$ by $- 1$ .

$\frac{- z^{2}}{- 1} = \frac{3}{- 1}$

Step 2.2

Simplify the left side.

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Step 2.2.1

Dividing two negative values results in a positive value.

$\frac{z^{2}}{1} = \frac{3}{- 1}$

Step 2.2.2

Divide $z^{2}$ by $1$ .

$z^{2} = \frac{3}{- 1}$

Step 2.3

Simplify the right side.

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Step 2.3.1

Divide $3$ by $- 1$ .

$z^{2} = - 3$

Step 3

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$z = \pm \sqrt{- 3}$

Step 4

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

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Step 4.1

Rewrite $- 3$ as $- 1 (3)$ .

$z = \pm \sqrt{- 1 (3)}$

Step 4.2

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$z = \pm \sqrt{- 1} \cdot \sqrt{3}$

Step 4.3

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

$z = \pm i \sqrt{3}$

Step 5

The complete solution is the result of both the positive and negative portions of the solution.

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Step 5.1

First, use the positive value of the $\pm$ to find the first solution.

$z = i \sqrt{3}$

Step 5.2

Next, use the negative value of the $\pm$ to find the second solution.

$z = - i \sqrt{3}$

Step 5.3

The complete solution is the result of both the positive and negative portions of the solution.

$z = i \sqrt{3}, - i \sqrt{3}$