Solve by Completing the Square 7b^2-14b-56=0

$7 b^{2} - 14 b - 56 = 0$

Step 1

Add $56$ to both sides of the equation.

$7 b^{2} - 14 b = 56$

Step 2

Divide each term in $7 b^{2} - 14 b = 56$ by $7$ and simplify.

Divide each term in $7 b^{2} - 14 b = 56$ by $7$ .

$\frac{7 b^{2}}{7} + \frac{- 14 b}{7} = \frac{56}{7}$

Simplify the left side.

Simplify each term.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 b^{2}}{7} + \frac{- 14 b}{7} = \frac{56}{7}$

Divide $b^{2}$ by $1$ .

$b^{2} + \frac{- 14 b}{7} = \frac{56}{7}$

Cancel the common factor of $- 14$ and $7$ .

Factor $7$ out of $- 14 b$ .

$b^{2} + \frac{7 (- 2 b)}{7} = \frac{56}{7}$

Cancel the common factors.

Factor $7$ out of $7$ .

$b^{2} + \frac{7 (- 2 b)}{7 (1)} = \frac{56}{7}$

Cancel the common factor.

$b^{2} + \frac{7 (- 2 b)}{7 \cdot 1} = \frac{56}{7}$

Rewrite the expression.

$b^{2} + \frac{- 2 b}{1} = \frac{56}{7}$

Divide $- 2 b$ by $1$ .

$b^{2} - 2 b = \frac{56}{7}$

Simplify the right side.

Divide $56$ by $7$ .

$b^{2} - 2 b = 8$

Step 3

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

${(\frac{b}{2})}^{2} = {(- 1)}^{2}$

Step 4

Add the term to each side of the equation.

$b^{2} - 2 b + {(- 1)}^{2} = 8 + {(- 1)}^{2}$

Step 5

Simplify the equation.

Simplify the left side.

Raise $- 1$ to the power of $2$ .

$b^{2} - 2 b + 1 = 8 + {(- 1)}^{2}$

Simplify the right side.

Simplify $8 + {(- 1)}^{2}$ .

Raise $- 1$ to the power of $2$ .

$b^{2} - 2 b + 1 = 8 + 1$

Add $8$ and $1$ .

$b^{2} - 2 b + 1 = 9$

Step 6

Factor the perfect trinomial square into ${(b - 1)}^{2}$ .

${(b - 1)}^{2} = 9$

Step 7

Solve the equation for $b$ .

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

$b - 1 = \pm \sqrt{9}$

Simplify $\pm \sqrt{9}$ .

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

$b - 1 = \pm \sqrt{3^{2}}$

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

$b - 1 = \pm 3$

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.

$b - 1 = 3$

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

Add $1$ to both sides of the equation.

$b = 3 + 1$

Add $3$ and $1$ .

$b = 4$

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

$b - 1 = - 3$

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

Add $1$ to both sides of the equation.

$b = - 3 + 1$

Add $- 3$ and $1$ .

$b = - 2$

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

$b = 4, - 2$

Solve by Completing the Square 7b^2-14b-56=0

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