Finite Math Examples

$2 = 2 + 2 x - 15 + x^{2}$

Step 1

Simplify the equation into a proper form to complete the square.

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Step 1.1

Simplify $2 + 2 x - 15 + x^{2}$ .

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Step 1.1.1

Subtract $15$ from $2$ .

$2 = 2 x - 13 + x^{2}$

Step 1.1.2

Move $- 13$ .

$2 = 2 x + x^{2} - 13$

Step 1.1.3

Reorder $2 x$ and $x^{2}$ .

$2 = x^{2} + 2 x - 13$

Step 1.2

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$x^{2} + 2 x - 13 = 2$

Step 1.3

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

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Step 1.3.1

Add $13$ to both sides of the equation.

$x^{2} + 2 x = 2 + 13$

Step 1.3.2

Add $2$ and $13$ .

$x^{2} + 2 x = 15$

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 $b$ .

${(\frac{b}{2})}^{2} = {(1)}^{2}$

Step 3

Add the term to each side of the equation.

$x^{2} + 2 x + {(1)}^{2} = 15 + {(1)}^{2}$

Step 4

Simplify the equation.

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Step 4.1

Simplify the left side.

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Step 4.1.1

One to any power is one.

$x^{2} + 2 x + 1 = 15 + {(1)}^{2}$

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify $15 + {(1)}^{2}$ .

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Step 4.2.1.1

One to any power is one.

$x^{2} + 2 x + 1 = 15 + 1$

Step 4.2.1.2

Add $15$ and $1$ .

$x^{2} + 2 x + 1 = 16$

Step 5

Factor the perfect trinomial square into ${(x + 1)}^{2}$ .

${(x + 1)}^{2} = 16$

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.

$x + 1 = \pm \sqrt{16}$

Step 6.2

Simplify $\pm \sqrt{16}$ .

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Step 6.2.1

Rewrite $16$ as $4^{2}$ .

$x + 1 = \pm \sqrt{4^{2}}$

Step 6.2.2

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

$x + 1 = \pm 4$

Step 6.3

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

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Step 6.3.1

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

$x + 1 = 4$

Step 6.3.2

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

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Step 6.3.2.1

Subtract $1$ from both sides of the equation.

$x = 4 - 1$

Step 6.3.2.2

Subtract $1$ from $4$ .

$x = 3$

Step 6.3.3

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

$x + 1 = - 4$

Step 6.3.4

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

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Step 6.3.4.1

Subtract $1$ from both sides of the equation.

$x = - 4 - 1$

Step 6.3.4.2

Subtract $1$ from $- 4$ .

$x = - 5$

Step 6.3.5

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

$x = 3, - 5$