Solve by Completing the Square x^2+52=-20x-3

$x^{2} + 52 = - 20 x - 3$

Step 1

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

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

Subtract $52$ from both sides of the equation.

$x^{2} = - 20 x - 3 - 52$

Subtract $52$ from $- 3$ .

$x^{2} = - 20 x - 55$

Add $20 x$ to both sides of the equation.

$x^{2} + 20 x = - 55$

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 $b$ .

${(\frac{b}{2})}^{2} = {(10)}^{2}$

Step 3

Add the term to each side of the equation.

$x^{2} + 20 x + {(10)}^{2} = - 55 + {(10)}^{2}$

Step 4

Simplify the equation.

Simplify the left side.

Raise $10$ to the power of $2$ .

$x^{2} + 20 x + 100 = - 55 + {(10)}^{2}$

Simplify the right side.

Simplify $- 55 + {(10)}^{2}$ .

Raise $10$ to the power of $2$ .

$x^{2} + 20 x + 100 = - 55 + 100$

Add $- 55$ and $100$ .

$x^{2} + 20 x + 100 = 45$

Step 5

Factor the perfect trinomial square into ${(x + 10)}^{2}$ .

${(x + 10)}^{2} = 45$

Step 6

Solve the equation for $x$ .

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

$x + 10 = \pm \sqrt{45}$

Simplify $\pm \sqrt{45}$ .

Rewrite $45$ as $3^{2} \cdot 5$ .

Factor $9$ out of $45$ .

$x + 10 = \pm \sqrt{9 (5)}$

Rewrite $9$ as $3^{2}$ .

$x + 10 = \pm \sqrt{3^{2} \cdot 5}$

Pull terms out from under the radical.

$x + 10 = \pm 3 \sqrt{5}$

Subtract $10$ from both sides of the equation.

$x = \pm 3 \sqrt{5} - 10$

Step 7

The result can be shown in multiple forms.

Exact Form:

$x = \pm 3 \sqrt{5} - 10$

Decimal Form:

$x = - 3.29179606 \dots, - 16.70820393 \dots$

Solve by Completing the Square x^2+52=-20x-3

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