Solve by Completing the Square -2x=x^2-6

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

Subtract $x^{2}$ from both sides of the equation.

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

Divide each term in $- 2 x - x^{2} = - 6$ by $- 1$ and simplify.

Divide each term in $- 2 x - x^{2} = - 6$ by $- 1$ .

$\frac{- 2 x}{- 1} + \frac{- x^{2}}{- 1} = \frac{- 6}{- 1}$

Simplify the left side.

Simplify each term.

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

$- 1 \cdot (- 2 x) + \frac{- x^{2}}{- 1} = \frac{- 6}{- 1}$

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

$- (- 2 x) + \frac{- x^{2}}{- 1} = \frac{- 6}{- 1}$

Multiply $- 2$ by $- 1$ .

$2 x + \frac{- x^{2}}{- 1} = \frac{- 6}{- 1}$

Dividing two negative values results in a positive value.

$2 x + \frac{x^{2}}{1} = \frac{- 6}{- 1}$

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

$2 x + x^{2} = \frac{- 6}{- 1}$

Simplify the right side.

Divide $- 6$ by $- 1$ .

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

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.

$x^{2} + 2 x + {(1)}^{2} = 6 + {(1)}^{2}$

Simplify the equation.

Simplify the left side.

One to any power is one.

$x^{2} + 2 x + 1 = 6 + {(1)}^{2}$

Simplify the right side.

Simplify $6 + {(1)}^{2}$ .

One to any power is one.

$x^{2} + 2 x + 1 = 6 + 1$

Add $6$ and $1$ .

$x^{2} + 2 x + 1 = 7$

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

${(x + 1)}^{2} = 7$

Solve the equation for $x$ .

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

$x + 1 = \pm \sqrt{7}$

Subtract $1$ from both sides of the equation.

$x = \pm \sqrt{7} - 1$

The result can be shown in multiple forms.

Exact Form:

$x = \pm \sqrt{7} - 1$

Decimal Form:

$x = 1.64575131 \dots, - 3.64575131 \dots$

Solve by Completing the Square -2x=x^2-6

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