Finite Math Examples

$x^{2} + 4 x + 24 = 0$

Step 1

Subtract $24$ from both sides of the equation.

$x^{2} + 4 x = - 24$

Step 2

To create a trinomial square on the left side of the equation, find a value that is equal to the square of half of $b$ .

${(\frac{b}{2})}^{2} = {(2)}^{2}$

Step 3

Add the term to each side of the equation.

$x^{2} + 4 x + {(2)}^{2} = - 24 + {(2)}^{2}$

Step 4

Simplify the equation.

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Step 4.1

Simplify the left side.

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Step 4.1.1

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

$x^{2} + 4 x + 4 = - 24 + {(2)}^{2}$

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify $- 24 + {(2)}^{2}$ .

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Step 4.2.1.1

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

$x^{2} + 4 x + 4 = - 24 + 4$

Step 4.2.1.2

Add $- 24$ and $4$ .

$x^{2} + 4 x + 4 = - 20$

Step 5

Factor the perfect trinomial square into ${(x + 2)}^{2}$ .

${(x + 2)}^{2} = - 20$

Step 6

Solve the equation for $x$ .

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Step 6.1

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

$x + 2 = \pm \sqrt{- 20}$

Step 6.2

Simplify $\pm \sqrt{- 20}$ .

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Step 6.2.1

Rewrite $- 20$ as $- 1 (20)$ .

$x + 2 = \pm \sqrt{- 1 (20)}$

Step 6.2.2

Rewrite $\sqrt{- 1 (20)}$ as $\sqrt{- 1} \cdot \sqrt{20}$ .

$x + 2 = \pm \sqrt{- 1} \cdot \sqrt{20}$

Step 6.2.3

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

$x + 2 = \pm i \cdot \sqrt{20}$

Step 6.2.4

Rewrite $20$ as $2^{2} \cdot 5$ .

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Step 6.2.4.1

Factor $4$ out of $20$ .

$x + 2 = \pm i \cdot \sqrt{4 (5)}$

Step 6.2.4.2

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

$x + 2 = \pm i \cdot \sqrt{2^{2} \cdot 5}$

Step 6.2.5

Pull terms out from under the radical.

$x + 2 = \pm i \cdot (2 \sqrt{5})$

Step 6.2.6

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

$x + 2 = \pm 2 i \sqrt{5}$

Step 6.3

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

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Step 6.3.1

Subtract $2$ from both sides of the equation.

$x = \pm 2 i \sqrt{5} - 2$

Step 6.3.2

Reorder $\pm 2 i \sqrt{5}$ and $- 2$ .

$x = - 2 \pm 2 i \sqrt{5}$