Solve by Completing the Square 0=2x^2-16x-18

$0 = 2 x^{2} - 16 x - 18$

Simplify the equation into a proper form to complete the square.

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$2 x^{2} - 16 x - 18 = 0$

Add $18$ to both sides of the equation.

$2 x^{2} - 16 x = 18$

Divide each term in $2 x^{2} - 16 x = 18$ by $2$ and simplify.

Divide each term in $2 x^{2} - 16 x = 18$ by $2$ .

$\frac{2 x^{2}}{2} + \frac{- 16 x}{2} = \frac{18}{2}$

Simplify the left side.

Simplify each term.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x^{2}}{2} + \frac{- 16 x}{2} = \frac{18}{2}$

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

$x^{2} + \frac{- 16 x}{2} = \frac{18}{2}$

Cancel the common factor of $- 16$ and $2$ .

Factor $2$ out of $- 16 x$ .

$x^{2} + \frac{2 (- 8 x)}{2} = \frac{18}{2}$

Cancel the common factors.

Factor $2$ out of $2$ .

$x^{2} + \frac{2 (- 8 x)}{2 (1)} = \frac{18}{2}$

Cancel the common factor.

$x^{2} + \frac{2 (- 8 x)}{2 \cdot 1} = \frac{18}{2}$

Rewrite the expression.

$x^{2} + \frac{- 8 x}{1} = \frac{18}{2}$

Divide $- 8 x$ by $1$ .

$x^{2} - 8 x = \frac{18}{2}$

Simplify the right side.

Divide $18$ by $2$ .

$x^{2} - 8 x = 9$

To create a trinomial square on the left side of the equation, find a value that is equal to the square of half of $b$ .

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

Add the term to each side of the equation.

$x^{2} - 8 x + {(- 4)}^{2} = 9 + {(- 4)}^{2}$

Simplify the equation.

Simplify the left side.

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

$x^{2} - 8 x + 16 = 9 + {(- 4)}^{2}$

Simplify the right side.

Simplify $9 + {(- 4)}^{2}$ .

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

$x^{2} - 8 x + 16 = 9 + 16$

Add $9$ and $16$ .

$x^{2} - 8 x + 16 = 25$

Factor the perfect trinomial square into ${(x - 4)}^{2}$ .

${(x - 4)}^{2} = 25$

Solve the equation for $x$ .

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

$x - 4 = \pm \sqrt{25}$

Simplify $\pm \sqrt{25}$ .

Rewrite $25$ as $5^{2}$ .

$x - 4 = \pm \sqrt{5^{2}}$

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

$x - 4 = \pm 5$

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.

$x - 4 = 5$

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

Add $4$ to both sides of the equation.

$x = 5 + 4$

Add $5$ and $4$ .

$x = 9$

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

$x - 4 = - 5$

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

Add $4$ to both sides of the equation.

$x = - 5 + 4$

Add $- 5$ and $4$ .

$x = - 1$

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

$x = 9, - 1$

Solve by Completing the Square 0=2x^2-16x-18

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