Solve by Completing the Square x^2+4x-19=5

$x^{2} + 4 x - 19 = 5$

Step 1

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

Add $19$ to both sides of the equation.

$x^{2} + 4 x = 5 + 19$

Add $5$ and $19$ .

$x^{2} + 4 x = 24$

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 $b$ .

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

Step 3

Add the term to each side of the equation.

$x^{2} + 4 x + {(2)}^{2} = 24 + {(2)}^{2}$

Step 4

Simplify the equation.

Simplify the left side.

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

$x^{2} + 4 x + 4 = 24 + {(2)}^{2}$

Simplify the right side.

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

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

$x^{2} + 4 x + 4 = 24 + 4$

Add $24$ and $4$ .

$x^{2} + 4 x + 4 = 28$

Step 5

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

${(x + 2)}^{2} = 28$

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.

$x + 2 = \pm \sqrt{28}$

Simplify $\pm \sqrt{28}$ .

Rewrite $28$ as $2^{2} \cdot 7$ .

Factor $4$ out of $28$ .

$x + 2 = \pm \sqrt{4 (7)}$

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

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

Pull terms out from under the radical.

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

Subtract $2$ from both sides of the equation.

$x = \pm 2 \sqrt{7} - 2$

Step 7

The result can be shown in multiple forms.

Exact Form:

$x = \pm 2 \sqrt{7} - 2$

Decimal Form:

$x = 3.29150262 \dots, - 7.29150262 \dots$

Solve by Completing the Square x^2+4x-19=5

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