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

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

Step 1

Subtract $6$ from both sides of the equation.

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

Step 2

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

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

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

Simplify the right side.

Divide $- 6$ by $2$ .

$x^{2} + \frac{5 x}{2} = - 3$

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 $b$ .

${(\frac{b}{2})}^{2} = {(\frac{5}{4})}^{2}$

Step 4

Add the term to each side of the equation.

$x^{2} + \frac{5 x}{2} + {(\frac{5}{4})}^{2} = - 3 + {(\frac{5}{4})}^{2}$

Step 5

Simplify the equation.

Simplify the left side.

Simplify each term.

Apply the product rule to $\frac{5}{4}$ .

$x^{2} + \frac{5 x}{2} + \frac{5^{2}}{4^{2}} = - 3 + {(\frac{5}{4})}^{2}$

Raise $5$ to the power of $2$ .

$x^{2} + \frac{5 x}{2} + \frac{25}{4^{2}} = - 3 + {(\frac{5}{4})}^{2}$

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

$x^{2} + \frac{5 x}{2} + \frac{25}{16} = - 3 + {(\frac{5}{4})}^{2}$

Simplify the right side.

Simplify $- 3 + {(\frac{5}{4})}^{2}$ .

Simplify each term.

Apply the product rule to $\frac{5}{4}$ .

$x^{2} + \frac{5 x}{2} + \frac{25}{16} = - 3 + \frac{5^{2}}{4^{2}}$

Raise $5$ to the power of $2$ .

$x^{2} + \frac{5 x}{2} + \frac{25}{16} = - 3 + \frac{25}{4^{2}}$

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

$x^{2} + \frac{5 x}{2} + \frac{25}{16} = - 3 + \frac{25}{16}$

To write $- 3$ as a fraction with a common denominator, multiply by $\frac{16}{16}$ .

$x^{2} + \frac{5 x}{2} + \frac{25}{16} = - 3 \cdot \frac{16}{16} + \frac{25}{16}$

Combine $- 3$ and $\frac{16}{16}$ .

$x^{2} + \frac{5 x}{2} + \frac{25}{16} = \frac{- 3 \cdot 16}{16} + \frac{25}{16}$

Combine the numerators over the common denominator.

$x^{2} + \frac{5 x}{2} + \frac{25}{16} = \frac{- 3 \cdot 16 + 25}{16}$

Simplify the numerator.

Multiply $- 3$ by $16$ .

$x^{2} + \frac{5 x}{2} + \frac{25}{16} = \frac{- 48 + 25}{16}$

Add $- 48$ and $25$ .

$x^{2} + \frac{5 x}{2} + \frac{25}{16} = \frac{- 23}{16}$

Move the negative in front of the fraction.

$x^{2} + \frac{5 x}{2} + \frac{25}{16} = - \frac{23}{16}$

Step 6

Factor the perfect trinomial square into ${(x + \frac{5}{4})}^{2}$ .

${(x + \frac{5}{4})}^{2} = - \frac{23}{16}$

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.

$x + \frac{5}{4} = \pm \sqrt{- \frac{23}{16}}$

Simplify $\pm \sqrt{- \frac{23}{16}}$ .

Rewrite $- \frac{23}{16}$ as ${(\frac{i}{4})}^{2} \cdot 23$ .

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

$x + \frac{5}{4} = \pm \sqrt{i^{2} \frac{23}{16}}$

Factor the perfect power $1^{2}$ out of $23$ .

$x + \frac{5}{4} = \pm \sqrt{i^{2} \frac{1^{2} \cdot 23}{16}}$

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

$x + \frac{5}{4} = \pm \sqrt{i^{2} \frac{1^{2} \cdot 23}{4^{2} \cdot 1}}$

Rearrange the fraction $\frac{1^{2} \cdot 23}{4^{2} \cdot 1}$ .

$x + \frac{5}{4} = \pm \sqrt{i^{2} ({(\frac{1}{4})}^{2} \cdot 23)}$

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

$x + \frac{5}{4} = \pm \sqrt{{(\frac{i}{4})}^{2} \cdot 23}$

Pull terms out from under the radical.

$x + \frac{5}{4} = \pm \frac{i}{4} \sqrt{23}$

Combine $\frac{i}{4}$ and $\sqrt{23}$ .

$x + \frac{5}{4} = \pm \frac{i \sqrt{23}}{4}$

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

Subtract $\frac{5}{4}$ from both sides of the equation.

$x = \pm \frac{i \sqrt{23}}{4} - \frac{5}{4}$

Reorder $\pm \frac{i \sqrt{23}}{4}$ and $- \frac{5}{4}$ .

$x = - \frac{5}{4} \pm \frac{i \sqrt{23}}{4}$

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

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