Solve by Completing the Square 2x^2=-7x-29

$2 x^{2} = - 7 x - 29$

Step 1

Add $7 x$ to both sides of the equation.

$2 x^{2} + 7 x = - 29$

Step 2

Divide each term in $2 x^{2} + 7 x = - 29$ by $2$ and simplify.

Divide each term in $2 x^{2} + 7 x = - 29$ by $2$ .

$\frac{2 x^{2}}{2} + \frac{7 x}{2} = \frac{- 29}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x^{2}}{2} + \frac{7 x}{2} = \frac{- 29}{2}$

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

$x^{2} + \frac{7 x}{2} = \frac{- 29}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x^{2} + \frac{7 x}{2} = - \frac{29}{2}$

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 $b$ .

${(\frac{b}{2})}^{2} = {(\frac{7}{4})}^{2}$

Step 4

Add the term to each side of the equation.

$x^{2} + \frac{7 x}{2} + {(\frac{7}{4})}^{2} = - \frac{29}{2} + {(\frac{7}{4})}^{2}$

Step 5

Simplify the equation.

Simplify the left side.

Simplify each term.

Apply the product rule to $\frac{7}{4}$ .

$x^{2} + \frac{7 x}{2} + \frac{7^{2}}{4^{2}} = - \frac{29}{2} + {(\frac{7}{4})}^{2}$

Raise $7$ to the power of $2$ .

$x^{2} + \frac{7 x}{2} + \frac{49}{4^{2}} = - \frac{29}{2} + {(\frac{7}{4})}^{2}$

Raise $4$ to the power of $2$ .

$x^{2} + \frac{7 x}{2} + \frac{49}{16} = - \frac{29}{2} + {(\frac{7}{4})}^{2}$

Simplify the right side.

Simplify $- \frac{29}{2} + {(\frac{7}{4})}^{2}$ .

Simplify each term.

Apply the product rule to $\frac{7}{4}$ .

$x^{2} + \frac{7 x}{2} + \frac{49}{16} = - \frac{29}{2} + \frac{7^{2}}{4^{2}}$

Raise $7$ to the power of $2$ .

$x^{2} + \frac{7 x}{2} + \frac{49}{16} = - \frac{29}{2} + \frac{49}{4^{2}}$

Raise $4$ to the power of $2$ .

$x^{2} + \frac{7 x}{2} + \frac{49}{16} = - \frac{29}{2} + \frac{49}{16}$

To write $- \frac{29}{2}$ as a fraction with a common denominator, multiply by $\frac{8}{8}$ .

$x^{2} + \frac{7 x}{2} + \frac{49}{16} = - \frac{29}{2} \cdot \frac{8}{8} + \frac{49}{16}$

Write each expression with a common denominator of $16$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{29}{2}$ by $\frac{8}{8}$ .

$x^{2} + \frac{7 x}{2} + \frac{49}{16} = - \frac{29 \cdot 8}{2 \cdot 8} + \frac{49}{16}$

Multiply $2$ by $8$ .

$x^{2} + \frac{7 x}{2} + \frac{49}{16} = - \frac{29 \cdot 8}{16} + \frac{49}{16}$

Combine the numerators over the common denominator.

$x^{2} + \frac{7 x}{2} + \frac{49}{16} = \frac{- 29 \cdot 8 + 49}{16}$

Simplify the numerator.

Multiply $- 29$ by $8$ .

$x^{2} + \frac{7 x}{2} + \frac{49}{16} = \frac{- 232 + 49}{16}$

Add $- 232$ and $49$ .

$x^{2} + \frac{7 x}{2} + \frac{49}{16} = \frac{- 183}{16}$

Move the negative in front of the fraction.

$x^{2} + \frac{7 x}{2} + \frac{49}{16} = - \frac{183}{16}$

Step 6

Factor the perfect trinomial square into ${(x + \frac{7}{4})}^{2}$ .

${(x + \frac{7}{4})}^{2} = - \frac{183}{16}$

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.

$x + \frac{7}{4} = \pm \sqrt{- \frac{183}{16}}$

Simplify $\pm \sqrt{- \frac{183}{16}}$ .

Rewrite $- \frac{183}{16}$ as ${(\frac{i}{4})}^{2} \cdot 183$ .

Rewrite $- 1$ as $i^{2}$ .

$x + \frac{7}{4} = \pm \sqrt{i^{2} \frac{183}{16}}$

Factor the perfect power $1^{2}$ out of $183$ .

$x + \frac{7}{4} = \pm \sqrt{i^{2} \frac{1^{2} \cdot 183}{16}}$

Factor the perfect power $4^{2}$ out of $16$ .

$x + \frac{7}{4} = \pm \sqrt{i^{2} \frac{1^{2} \cdot 183}{4^{2} \cdot 1}}$

Rearrange the fraction $\frac{1^{2} \cdot 183}{4^{2} \cdot 1}$ .

$x + \frac{7}{4} = \pm \sqrt{i^{2} ({(\frac{1}{4})}^{2} \cdot 183)}$

Rewrite $i^{2} {(\frac{1}{4})}^{2}$ as ${(\frac{i}{4})}^{2}$ .

$x + \frac{7}{4} = \pm \sqrt{{(\frac{i}{4})}^{2} \cdot 183}$

Pull terms out from under the radical.

$x + \frac{7}{4} = \pm \frac{i}{4} \sqrt{183}$

Combine $\frac{i}{4}$ and $\sqrt{183}$ .

$x + \frac{7}{4} = \pm \frac{i \sqrt{183}}{4}$

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

Subtract $\frac{7}{4}$ from both sides of the equation.

$x = \pm \frac{i \sqrt{183}}{4} - \frac{7}{4}$

Reorder $\pm \frac{i \sqrt{183}}{4}$ and $- \frac{7}{4}$ .

$x = - \frac{7}{4} \pm \frac{i \sqrt{183}}{4}$

Solve by Completing the Square 2x^2=-7x-29

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