Solve by Completing the Square 2x^2-4x-16=0

$2 x^{2} - 4 x - 16 = 0$

Add $16$ to both sides of the equation.

$2 x^{2} - 4 x = 16$

Divide each term in $2 x^{2} - 4 x = 16$ by $2$ and simplify.

Divide each term in $2 x^{2} - 4 x = 16$ by $2$ .

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

Simplify the left side.

Simplify each term.

Cancel the common factor of $2$ .

Cancel the common factor.

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

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

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

Cancel the common factor of $- 4$ and $2$ .

Factor $2$ out of $- 4 x$ .

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

Cancel the common factors.

Factor $2$ out of $2$ .

$x^{2} + \frac{2 (- 2 x)}{2 (1)} = \frac{16}{2}$

Cancel the common factor.

$x^{2} + \frac{2 (- 2 x)}{2 \cdot 1} = \frac{16}{2}$

Rewrite the expression.

$x^{2} + \frac{- 2 x}{1} = \frac{16}{2}$

Divide $- 2 x$ by $1$ .

$x^{2} - 2 x = \frac{16}{2}$

Simplify the right side.

Divide $16$ by $2$ .

$x^{2} - 2 x = 8$

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.

$x^{2} - 2 x + {(- 1)}^{2} = 8 + {(- 1)}^{2}$

Simplify the equation.

Simplify the left side.

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

$x^{2} - 2 x + 1 = 8 + {(- 1)}^{2}$

Simplify the right side.

Simplify $8 + {(- 1)}^{2}$ .

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

$x^{2} - 2 x + 1 = 8 + 1$

Add $8$ and $1$ .

$x^{2} - 2 x + 1 = 9$

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

${(x - 1)}^{2} = 9$

Solve the equation for $x$ .

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

$x - 1 = \pm \sqrt{9}$

Simplify $\pm \sqrt{9}$ .

Rewrite $9$ as $3^{2}$ .

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

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

$x - 1 = \pm 3$

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.

$x - 1 = 3$

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

Add $1$ to both sides of the equation.

$x = 3 + 1$

Add $3$ and $1$ .

$x = 4$

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

$x - 1 = - 3$

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

Add $1$ to both sides of the equation.

$x = - 3 + 1$

Add $- 3$ and $1$ .

$x = - 2$

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

$x = 4, - 2$

Solve by Completing the Square 2x^2-4x-16=0

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