Solve by Completing the Square x^2+14x-51=0

$x^{2} + 14 x - 51 = 0$

Step 1

Add $51$ to both sides of the equation.

$x^{2} + 14 x = 51$

Step 2

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

${(\frac{b}{2})}^{2} = {(7)}^{2}$

Step 3

Add the term to each side of the equation.

$x^{2} + 14 x + {(7)}^{2} = 51 + {(7)}^{2}$

Step 4

Simplify the equation.

Simplify the left side.

Raise $7$ to the power of $2$ .

$x^{2} + 14 x + 49 = 51 + {(7)}^{2}$

Simplify the right side.

Simplify $51 + {(7)}^{2}$ .

Raise $7$ to the power of $2$ .

$x^{2} + 14 x + 49 = 51 + 49$

Add $51$ and $49$ .

$x^{2} + 14 x + 49 = 100$

Step 5

Factor the perfect trinomial square into ${(x + 7)}^{2}$ .

${(x + 7)}^{2} = 100$

Step 6

Solve the equation for $x$ .

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

$x + 7 = \pm \sqrt{100}$

Simplify $\pm \sqrt{100}$ .

Rewrite $100$ as $10^{2}$ .

$x + 7 = \pm \sqrt{10^{2}}$

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

$x + 7 = \pm 10$

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.

$x + 7 = 10$

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

Subtract $7$ from both sides of the equation.

$x = 10 - 7$

Subtract $7$ from $10$ .

$x = 3$

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

$x + 7 = - 10$

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

Subtract $7$ from both sides of the equation.

$x = - 10 - 7$

Subtract $7$ from $- 10$ .

$x = - 17$

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

$x = 3, - 17$

Solve by Completing the Square x^2+14x-51=0

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