Solve by Completing the Square x^2+16x-145=10x-7

$x^{2} + 16 x - 145 = 10 x - 7$

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

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

Add $145$ to both sides of the equation.

$x^{2} + 16 x = 10 x - 7 + 145$

Add $- 7$ and $145$ .

$x^{2} + 16 x = 10 x + 138$

Move all terms containing $x$ to the left side of the equation.

Subtract $10 x$ from both sides of the equation.

$x^{2} + 16 x - 10 x = 138$

Subtract $10 x$ from $16 x$ .

$x^{2} + 6 x = 138$

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

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

Add the term to each side of the equation.

$x^{2} + 6 x + {(3)}^{2} = 138 + {(3)}^{2}$

Simplify the equation.

Simplify the left side.

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

$x^{2} + 6 x + 9 = 138 + {(3)}^{2}$

Simplify the right side.

Simplify $138 + {(3)}^{2}$ .

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

$x^{2} + 6 x + 9 = 138 + 9$

Add $138$ and $9$ .

$x^{2} + 6 x + 9 = 147$

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

${(x + 3)}^{2} = 147$

Solve the equation for $x$ .

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

$x + 3 = \pm \sqrt{147}$

Simplify $\pm \sqrt{147}$ .

Rewrite $147$ as $7^{2} \cdot 3$ .

Factor $49$ out of $147$ .

$x + 3 = \pm \sqrt{49 (3)}$

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

$x + 3 = \pm \sqrt{7^{2} \cdot 3}$

Pull terms out from under the radical.

$x + 3 = \pm 7 \sqrt{3}$

Subtract $3$ from both sides of the equation.

$x = \pm 7 \sqrt{3} - 3$

The result can be shown in multiple forms.

Exact Form:

$x = \pm 7 \sqrt{3} - 3$

Decimal Form:

$x = 9.12435565 \dots, - 15.12435565 \dots$

Solve by Completing the Square x^2+16x-145=10x-7

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