Solve by Completing the Square x^2+10x=24

$x^{2} + 10 x = 24$

Step 1

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

${(\frac{b}{2})}^{2} = {(5)}^{2}$

Step 2

Add the term to each side of the equation.

$x^{2} + 10 x + {(5)}^{2} = 24 + {(5)}^{2}$

Step 3

Simplify the equation.

Simplify the left side.

Raise $5$ to the power of $2$ .

$x^{2} + 10 x + 25 = 24 + {(5)}^{2}$

Simplify the right side.

Simplify $24 + {(5)}^{2}$ .

Raise $5$ to the power of $2$ .

$x^{2} + 10 x + 25 = 24 + 25$

Add $24$ and $25$ .

$x^{2} + 10 x + 25 = 49$

Step 4

Factor the perfect trinomial square into ${(x + 5)}^{2}$ .

${(x + 5)}^{2} = 49$

Step 5

Solve the equation for $x$ .

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

$x + 5 = \pm \sqrt{49}$

Simplify $\pm \sqrt{49}$ .

Rewrite $49$ as $7^{2}$ .

$x + 5 = \pm \sqrt{7^{2}}$

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

$x + 5 = \pm 7$

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.

$x + 5 = 7$

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

Subtract $5$ from both sides of the equation.

$x = 7 - 5$

Subtract $5$ from $7$ .

$x = 2$

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

$x + 5 = - 7$

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

Subtract $5$ from both sides of the equation.

$x = - 7 - 5$

Subtract $5$ from $- 7$ .

$x = - 12$

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

$x = 2, - 12$

Solve by Completing the Square x^2+10x=24

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