Solve by Completing the Square 6x^2+13x-5=0

$6 x^{2} + 13 x - 5 = 0$

Step 1

Add $5$ to both sides of the equation.

$6 x^{2} + 13 x = 5$

Step 2

Divide each term in $6 x^{2} + 13 x = 5$ by $6$ and simplify.

Divide each term in $6 x^{2} + 13 x = 5$ by $6$ .

$\frac{6 x^{2}}{6} + \frac{13 x}{6} = \frac{5}{6}$

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6 x^{2}}{6} + \frac{13 x}{6} = \frac{5}{6}$

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

$x^{2} + \frac{13 x}{6} = \frac{5}{6}$

Step 3

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

${(\frac{b}{2})}^{2} = {(\frac{13}{12})}^{2}$

Step 4

Add the term to each side of the equation.

$x^{2} + \frac{13 x}{6} + {(\frac{13}{12})}^{2} = \frac{5}{6} + {(\frac{13}{12})}^{2}$

Step 5

Simplify the equation.

Simplify the left side.

Simplify each term.

Apply the product rule to $\frac{13}{12}$ .

$x^{2} + \frac{13 x}{6} + \frac{13^{2}}{12^{2}} = \frac{5}{6} + {(\frac{13}{12})}^{2}$

Raise $13$ to the power of $2$ .

$x^{2} + \frac{13 x}{6} + \frac{169}{12^{2}} = \frac{5}{6} + {(\frac{13}{12})}^{2}$

Raise $12$ to the power of $2$ .

$x^{2} + \frac{13 x}{6} + \frac{169}{144} = \frac{5}{6} + {(\frac{13}{12})}^{2}$

Simplify the right side.

Simplify $\frac{5}{6} + {(\frac{13}{12})}^{2}$ .

Simplify each term.

Apply the product rule to $\frac{13}{12}$ .

$x^{2} + \frac{13 x}{6} + \frac{169}{144} = \frac{5}{6} + \frac{13^{2}}{12^{2}}$

Raise $13$ to the power of $2$ .

$x^{2} + \frac{13 x}{6} + \frac{169}{144} = \frac{5}{6} + \frac{169}{12^{2}}$

Raise $12$ to the power of $2$ .

$x^{2} + \frac{13 x}{6} + \frac{169}{144} = \frac{5}{6} + \frac{169}{144}$

To write $\frac{5}{6}$ as a fraction with a common denominator, multiply by $\frac{24}{24}$ .

$x^{2} + \frac{13 x}{6} + \frac{169}{144} = \frac{5}{6} \cdot \frac{24}{24} + \frac{169}{144}$

Write each expression with a common denominator of $144$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{5}{6}$ by $\frac{24}{24}$ .

$x^{2} + \frac{13 x}{6} + \frac{169}{144} = \frac{5 \cdot 24}{6 \cdot 24} + \frac{169}{144}$

Multiply $6$ by $24$ .

$x^{2} + \frac{13 x}{6} + \frac{169}{144} = \frac{5 \cdot 24}{144} + \frac{169}{144}$

Combine the numerators over the common denominator.

$x^{2} + \frac{13 x}{6} + \frac{169}{144} = \frac{5 \cdot 24 + 169}{144}$

Simplify the numerator.

Multiply $5$ by $24$ .

$x^{2} + \frac{13 x}{6} + \frac{169}{144} = \frac{120 + 169}{144}$

Add $120$ and $169$ .

$x^{2} + \frac{13 x}{6} + \frac{169}{144} = \frac{289}{144}$

Step 6

Factor the perfect trinomial square into ${(x + \frac{13}{12})}^{2}$ .

${(x + \frac{13}{12})}^{2} = \frac{289}{144}$

Step 7

Solve the equation for $x$ .

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

$x + \frac{13}{12} = \pm \sqrt{\frac{289}{144}}$

Simplify $\pm \sqrt{\frac{289}{144}}$ .

Rewrite $\sqrt{\frac{289}{144}}$ as $\frac{\sqrt{289}}{\sqrt{144}}$ .

$x + \frac{13}{12} = \pm \frac{\sqrt{289}}{\sqrt{144}}$

Simplify the numerator.

Rewrite $289$ as $17^{2}$ .

$x + \frac{13}{12} = \pm \frac{\sqrt{17^{2}}}{\sqrt{144}}$

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

$x + \frac{13}{12} = \pm \frac{17}{\sqrt{144}}$

Simplify the denominator.

Rewrite $144$ as $12^{2}$ .

$x + \frac{13}{12} = \pm \frac{17}{\sqrt{12^{2}}}$

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

$x + \frac{13}{12} = \pm \frac{17}{12}$

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

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

$x + \frac{13}{12} = \frac{17}{12}$

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

Subtract $\frac{13}{12}$ from both sides of the equation.

$x = \frac{17}{12} - \frac{13}{12}$

Combine the numerators over the common denominator.

$x = \frac{17 - 13}{12}$

Subtract $13$ from $17$ .

$x = \frac{4}{12}$

Cancel the common factor of $4$ and $12$ .

Factor $4$ out of $4$ .

$x = \frac{4 (1)}{12}$

Cancel the common factors.

Factor $4$ out of $12$ .

$x = \frac{4 \cdot 1}{4 \cdot 3}$

Cancel the common factor.

$x = \frac{4 \cdot 1}{4 \cdot 3}$

Rewrite the expression.

$x = \frac{1}{3}$

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

$x + \frac{13}{12} = - \frac{17}{12}$

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

Subtract $\frac{13}{12}$ from both sides of the equation.

$x = - \frac{17}{12} - \frac{13}{12}$

Combine the numerators over the common denominator.

$x = \frac{- 17 - 13}{12}$

Subtract $13$ from $- 17$ .

$x = \frac{- 30}{12}$

Cancel the common factor of $- 30$ and $12$ .

Factor $6$ out of $- 30$ .

$x = \frac{6 (- 5)}{12}$

Cancel the common factors.

Factor $6$ out of $12$ .

$x = \frac{6 \cdot - 5}{6 \cdot 2}$

Cancel the common factor.

$x = \frac{6 \cdot - 5}{6 \cdot 2}$

Rewrite the expression.

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

Move the negative in front of the fraction.

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

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

$x = \frac{1}{3}, - \frac{5}{2}$

Solve by Completing the Square 6x^2+13x-5=0

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