Finite Math Examples

$(z + 2) \cdot (z - 3) = 1$

Step 1

Move all terms to the left side of the equation and simplify.

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Step 1.1

Simplify the left side.

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Step 1.1.1

Simplify $(z + 2) \cdot (z - 3)$ .

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Step 1.1.1.1

Expand $(z + 2) (z - 3)$ using the FOIL Method.

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Step 1.1.1.1.1

Apply the distributive property.

$z (z - 3) + 2 (z - 3) = 1$

Step 1.1.1.1.2

Apply the distributive property.

$z \cdot z + z \cdot - 3 + 2 (z - 3) = 1$

Step 1.1.1.1.3

Apply the distributive property.

$z \cdot z + z \cdot - 3 + 2 z + 2 \cdot - 3 = 1$

Step 1.1.1.2

Simplify and combine like terms.

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Step 1.1.1.2.1

Simplify each term.

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Step 1.1.1.2.1.1

Multiply $z$ by $z$ .

$z^{2} + z \cdot - 3 + 2 z + 2 \cdot - 3 = 1$

Step 1.1.1.2.1.2

Move $- 3$ to the left of $z$ .

$z^{2} - 3 \cdot z + 2 z + 2 \cdot - 3 = 1$

Step 1.1.1.2.1.3

Multiply $2$ by $- 3$ .

$z^{2} - 3 z + 2 z - 6 = 1$

Step 1.1.1.2.2

Add $- 3 z$ and $2 z$ .

$z^{2} - z - 6 = 1$

Step 1.2

Subtract $1$ from both sides of the equation.

$z^{2} - z - 6 - 1 = 0$

Step 1.3

Subtract $1$ from $- 6$ .

$z^{2} - z - 7 = 0$

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 $a = 1$ , $b = - 1$ , and $c = - 7$ into the quadratic formula and solve for $z$ .

$\frac{1 \pm \sqrt{{(- 1)}^{2} - 4 \cdot (1 \cdot - 7)}}{2 \cdot 1}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Raise $- 1$ to the power of $2$ .

$z = \frac{1 \pm \sqrt{1 - 4 \cdot 1 \cdot - 7}}{2 \cdot 1}$

Step 4.1.2

Multiply $- 4 \cdot 1 \cdot - 7$ .

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Step 4.1.2.1

Multiply $- 4$ by $1$ .

$z = \frac{1 \pm \sqrt{1 - 4 \cdot - 7}}{2 \cdot 1}$

Step 4.1.2.2

Multiply $- 4$ by $- 7$ .

$z = \frac{1 \pm \sqrt{1 + 28}}{2 \cdot 1}$

Step 4.1.3

Add $1$ and $28$ .

$z = \frac{1 \pm \sqrt{29}}{2 \cdot 1}$

Step 4.2

Multiply $2$ by $1$ .

$z = \frac{1 \pm \sqrt{29}}{2}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$z = \frac{1 \pm \sqrt{29}}{2}$

Decimal Form:

$z = 3.19258240 \dots, - 2.19258240 \dots$