Solve by Completing the Square 2x^2+12x-14=0

$2 x^{2} + 12 x - 14 = 0$

Step 1

Add $14$ to both sides of the equation.

$2 x^{2} + 12 x = 14$

Step 2

Divide each term in $2 x^{2} + 12 x = 14$ by $2$ and simplify.

Divide each term in $2 x^{2} + 12 x = 14$ by $2$ .

$\frac{2 x^{2}}{2} + \frac{12 x}{2} = \frac{14}{2}$

Simplify the left side.

Simplify each term.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x^{2}}{2} + \frac{12 x}{2} = \frac{14}{2}$

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

$x^{2} + \frac{12 x}{2} = \frac{14}{2}$

Cancel the common factor of $12$ and $2$ .

Factor $2$ out of $12 x$ .

$x^{2} + \frac{2 (6 x)}{2} = \frac{14}{2}$

Cancel the common factors.

Factor $2$ out of $2$ .

$x^{2} + \frac{2 (6 x)}{2 (1)} = \frac{14}{2}$

Cancel the common factor.

$x^{2} + \frac{2 (6 x)}{2 \cdot 1} = \frac{14}{2}$

Rewrite the expression.

$x^{2} + \frac{6 x}{1} = \frac{14}{2}$

Divide $6 x$ by $1$ .

$x^{2} + 6 x = \frac{14}{2}$

Simplify the right side.

Divide $14$ by $2$ .

$x^{2} + 6 x = 7$

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 $b$ .

${(\frac{b}{2})}^{2} = {(3)}^{2}$

Step 4

Add the term to each side of the equation.

$x^{2} + 6 x + {(3)}^{2} = 7 + {(3)}^{2}$

Step 5

Simplify the equation.

Simplify the left side.

Raise $3$ to the power of $2$ .

$x^{2} + 6 x + 9 = 7 + {(3)}^{2}$

Simplify the right side.

Simplify $7 + {(3)}^{2}$ .

Raise $3$ to the power of $2$ .

$x^{2} + 6 x + 9 = 7 + 9$

Add $7$ and $9$ .

$x^{2} + 6 x + 9 = 16$

Step 6

Factor the perfect trinomial square into ${(x + 3)}^{2}$ .

${(x + 3)}^{2} = 16$

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.

$x + 3 = \pm \sqrt{16}$

Simplify $\pm \sqrt{16}$ .

Rewrite $16$ as $4^{2}$ .

$x + 3 = \pm \sqrt{4^{2}}$

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

$x + 3 = \pm 4$

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.

$x + 3 = 4$

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

Subtract $3$ from both sides of the equation.

$x = 4 - 3$

Subtract $3$ from $4$ .

$x = 1$

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

$x + 3 = - 4$

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

Subtract $3$ from both sides of the equation.

$x = - 4 - 3$

Subtract $3$ from $- 4$ .

$x = - 7$

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

$x = 1, - 7$

Solve by Completing the Square 2x^2+12x-14=0

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