Solve by Completing the Square 6x^2-12x-18=0

$6 x^{2} - 12 x - 18 = 0$

Step 1

Add $18$ to both sides of the equation.

$6 x^{2} - 12 x = 18$

Step 2

Divide each term in $6 x^{2} - 12 x = 18$ by $6$ and simplify.

Divide each term in $6 x^{2} - 12 x = 18$ by $6$ .

$\frac{6 x^{2}}{6} + \frac{- 12 x}{6} = \frac{18}{6}$

Simplify the left side.

Simplify each term.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6 x^{2}}{6} + \frac{- 12 x}{6} = \frac{18}{6}$

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

$x^{2} + \frac{- 12 x}{6} = \frac{18}{6}$

Cancel the common factor of $- 12$ and $6$ .

Factor $6$ out of $- 12 x$ .

$x^{2} + \frac{6 (- 2 x)}{6} = \frac{18}{6}$

Cancel the common factors.

Factor $6$ out of $6$ .

$x^{2} + \frac{6 (- 2 x)}{6 (1)} = \frac{18}{6}$

Cancel the common factor.

$x^{2} + \frac{6 (- 2 x)}{6 \cdot 1} = \frac{18}{6}$

Rewrite the expression.

$x^{2} + \frac{- 2 x}{1} = \frac{18}{6}$

Divide $- 2 x$ by $1$ .

$x^{2} - 2 x = \frac{18}{6}$

Simplify the right side.

Divide $18$ by $6$ .

$x^{2} - 2 x = 3$

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 $b$ .

${(\frac{b}{2})}^{2} = {(- 1)}^{2}$

Step 4

Add the term to each side of the equation.

$x^{2} - 2 x + {(- 1)}^{2} = 3 + {(- 1)}^{2}$

Step 5

Simplify the equation.

Simplify the left side.

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

$x^{2} - 2 x + 1 = 3 + {(- 1)}^{2}$

Simplify the right side.

Simplify $3 + {(- 1)}^{2}$ .

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

$x^{2} - 2 x + 1 = 3 + 1$

Add $3$ and $1$ .

$x^{2} - 2 x + 1 = 4$

Step 6

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

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

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.

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

Simplify $\pm \sqrt{4}$ .

Rewrite $4$ as $2^{2}$ .

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

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

$x - 1 = \pm 2$

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.

$x - 1 = 2$

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

Add $1$ to both sides of the equation.

$x = 2 + 1$

Add $2$ and $1$ .

$x = 3$

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

$x - 1 = - 2$

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

Add $1$ to both sides of the equation.

$x = - 2 + 1$

Add $- 2$ and $1$ .

$x = - 1$

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

$x = 3, - 1$

Solve by Completing the Square 6x^2-12x-18=0

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