Solve by Completing the Square -4x^2+8x=7

$- 4 x^{2} + 8 x = 7$

Step 1

Divide each term in $- 4 x^{2} + 8 x = 7$ by $- 4$ and simplify.

Divide each term in $- 4 x^{2} + 8 x = 7$ by $- 4$ .

$\frac{- 4 x^{2}}{- 4} + \frac{8 x}{- 4} = \frac{7}{- 4}$

Simplify the left side.

Simplify each term.

Cancel the common factor of $- 4$ .

Cancel the common factor.

$\frac{- 4 x^{2}}{- 4} + \frac{8 x}{- 4} = \frac{7}{- 4}$

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

$x^{2} + \frac{8 x}{- 4} = \frac{7}{- 4}$

Cancel the common factor of $8$ and $- 4$ .

Factor $4$ out of $8 x$ .

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

Move the negative one from the denominator of $\frac{2 x}{- 1}$ .

$x^{2} - 1 \cdot (2 x) = \frac{7}{- 4}$

Rewrite $- 1 \cdot (2 x)$ as $- (2 x)$ .

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

Multiply $2$ by $- 1$ .

$x^{2} - 2 x = \frac{7}{- 4}$

Simplify the right side.

Move the negative in front of the fraction.

$x^{2} - 2 x = - \frac{7}{4}$

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.

$x^{2} - 2 x + {(- 1)}^{2} = - \frac{7}{4} + {(- 1)}^{2}$

Step 4

Simplify the equation.

Simplify the left side.

Raise $- 1$ to the power of $2$ .

$x^{2} - 2 x + 1 = - \frac{7}{4} + {(- 1)}^{2}$

Simplify the right side.

Simplify $- \frac{7}{4} + {(- 1)}^{2}$ .

Raise $- 1$ to the power of $2$ .

$x^{2} - 2 x + 1 = - \frac{7}{4} + 1$

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

$x^{2} - 2 x + 1 = - \frac{7}{4} + \frac{4}{4}$

Combine the numerators over the common denominator.

$x^{2} - 2 x + 1 = \frac{- 7 + 4}{4}$

Add $- 7$ and $4$ .

$x^{2} - 2 x + 1 = \frac{- 3}{4}$

Move the negative in front of the fraction.

$x^{2} - 2 x + 1 = - \frac{3}{4}$

Step 5

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

${(x - 1)}^{2} = - \frac{3}{4}$

Step 6

Solve the equation for $x$ .

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

$x - 1 = \pm \sqrt{- \frac{3}{4}}$

Simplify $\pm \sqrt{- \frac{3}{4}}$ .

Rewrite $- \frac{3}{4}$ as ${(\frac{i}{2})}^{2} \cdot 3$ .

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

$x - 1 = \pm \sqrt{i^{2} \frac{3}{4}}$

Factor the perfect power $1^{2}$ out of $3$ .

$x - 1 = \pm \sqrt{i^{2} \frac{1^{2} \cdot 3}{4}}$

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

$x - 1 = \pm \sqrt{i^{2} \frac{1^{2} \cdot 3}{2^{2} \cdot 1}}$

Rearrange the fraction $\frac{1^{2} \cdot 3}{2^{2} \cdot 1}$ .

$x - 1 = \pm \sqrt{i^{2} ({(\frac{1}{2})}^{2} \cdot 3)}$

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

$x - 1 = \pm \sqrt{{(\frac{i}{2})}^{2} \cdot 3}$

Pull terms out from under the radical.

$x - 1 = \pm \frac{i}{2} \sqrt{3}$

Combine $\frac{i}{2}$ and $\sqrt{3}$ .

$x - 1 = \pm \frac{i \sqrt{3}}{2}$

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

Add $1$ to both sides of the equation.

$x = \pm \frac{i \sqrt{3}}{2} + 1$

Reorder $\pm \frac{i \sqrt{3}}{2}$ and $1$ .

$x = 1 \pm \frac{i \sqrt{3}}{2}$

Solve by Completing the Square -4x^2+8x=7

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