Solve by Completing the Square 8x^2+16x=42

$8 x^{2} + 16 x = 42$

Divide each term by $8$ and simplify.

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Divide each term in $8 x^{2} + 16 x = 42$ by $8$ .

$\frac{8 x^{2}}{8} + \frac{16 x}{8} = \frac{42}{8}$

Simplify each term.

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Cancel the common factor of $8$ .

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Cancel the common factor.

$\frac{8 x^{2}}{8} + \frac{16 x}{8} = \frac{42}{8}$

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

$x^{2} + \frac{16 x}{8} = \frac{42}{8}$

Cancel the common factor of $16$ and $8$ .

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Factor $8$ out of $16 x$ .

$x^{2} + \frac{8 (2 x)}{8} = \frac{42}{8}$

Cancel the common factors.

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Factor $8$ out of $8$ .

$x^{2} + \frac{8 (2 x)}{8 (1)} = \frac{42}{8}$

Cancel the common factor.

$x^{2} + \frac{8 (2 x)}{8 \cdot 1} = \frac{42}{8}$

Rewrite the expression.

$x^{2} + \frac{2 x}{1} = \frac{42}{8}$

Divide $2 x$ by $1$ .

$x^{2} + 2 x = \frac{42}{8}$

Cancel the common factor of $42$ and $8$ .

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Factor $2$ out of $42$ .

$x^{2} + 2 x = \frac{2 (21)}{8}$

Cancel the common factors.

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Factor $2$ out of $8$ .

$x^{2} + 2 x = \frac{2 \cdot 21}{2 \cdot 4}$

Cancel the common factor.

$x^{2} + 2 x = \frac{2 \cdot 21}{2 \cdot 4}$

Rewrite the expression.

$x^{2} + 2 x = \frac{21}{4}$

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

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

Add the term to each side of the equation.

$x^{2} + 2 x + {(1)}^{2} = \frac{21}{4} + {(1)}^{2}$

Simplify the equation.

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One to any power is one.

$x^{2} + 2 x + 1 = \frac{21}{4} + {(1)}^{2}$

Simplify $\frac{21}{4} + {(1)}^{2}$ .

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One to any power is one.

$x^{2} + 2 x + 1 = \frac{21}{4} + 1$

Write $1$ as a fraction with a common denominator.

$x^{2} + 2 x + 1 = \frac{21}{4} + \frac{4}{4}$

Combine the numerators over the common denominator.

$x^{2} + 2 x + 1 = \frac{21 + 4}{4}$

Add $21$ and $4$ .

$x^{2} + 2 x + 1 = \frac{25}{4}$

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

${(x + 1)}^{2} = \frac{25}{4}$

Solve the equation for $x$ .

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Take the square root of each side of the equation to set up the solution for $x$

${(x + 1)}^{2 \cdot \frac{1}{2}} = \pm \sqrt{\frac{25}{4}}$

Remove the perfect root factor $x + 1$ under the radical to solve for $x$ .

$x + 1 = \pm \sqrt{\frac{25}{4}}$

Simplify the right side of the equation.

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Rewrite $\sqrt{\frac{25}{4}}$ as $\frac{\sqrt{25}}{\sqrt{4}}$ .

$x + 1 = \pm \frac{\sqrt{25}}{\sqrt{4}}$

Simplify the numerator.

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Rewrite $25$ as $5^{2}$ .

$x + 1 = \pm \frac{\sqrt{5^{2}}}{\sqrt{4}}$

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

$x + 1 = \pm \frac{5}{\sqrt{4}}$

Simplify the denominator.

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

$x + 1 = \pm \frac{5}{\sqrt{2^{2}}}$

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

$x + 1 = \pm \frac{5}{2}$

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

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

$x + 1 = \frac{5}{2}$

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

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Subtract $1$ from both sides of the equation.

$x = \frac{5}{2} - 1$

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$x = \frac{5}{2} - 1 \cdot \frac{2}{2}$

Combine $- 1$ and $\frac{2}{2}$ .

$x = \frac{5}{2} + \frac{- 1 \cdot 2}{2}$

Combine the numerators over the common denominator.

$x = \frac{5 - 1 \cdot 2}{2}$

Simplify the numerator.

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Multiply $- 1$ by $2$ .

$x = \frac{5 - 2}{2}$

Subtract $2$ from $5$ .

$x = \frac{3}{2}$

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

$x + 1 = - \frac{5}{2}$

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

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Subtract $1$ from both sides of the equation.

$x = - \frac{5}{2} - 1$

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$x = - \frac{5}{2} - 1 \cdot \frac{2}{2}$

Combine $- 1$ and $\frac{2}{2}$ .

$x = - \frac{5}{2} + \frac{- 1 \cdot 2}{2}$

Combine the numerators over the common denominator.

$x = \frac{- 5 - 1 \cdot 2}{2}$

Simplify the numerator.

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Multiply $- 1$ by $2$ .

$x = \frac{- 5 - 2}{2}$

Subtract $2$ from $- 5$ .

$x = \frac{- 7}{2}$

Move the negative in front of the fraction.

$x = - \frac{7}{2}$

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

$x = \frac{3}{2}, - \frac{7}{2}$

Solve by Completing the Square 8x^2+16x=42

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