Solve by Completing the Square y^2-12y=-35

$y^{2} - 12 y = - 35$

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 $b$ .

${(\frac{b}{2})}^{2} = {(- 6)}^{2}$

Step 2

Add the term to each side of the equation.

$y^{2} - 12 y + {(- 6)}^{2} = - 35 + {(- 6)}^{2}$

Step 3

Simplify the equation.

Simplify the left side.

Raise $- 6$ to the power of $2$ .

$y^{2} - 12 y + 36 = - 35 + {(- 6)}^{2}$

Simplify the right side.

Simplify $- 35 + {(- 6)}^{2}$ .

Raise $- 6$ to the power of $2$ .

$y^{2} - 12 y + 36 = - 35 + 36$

Add $- 35$ and $36$ .

$y^{2} - 12 y + 36 = 1$

Step 4

Factor the perfect trinomial square into ${(y - 6)}^{2}$ .

${(y - 6)}^{2} = 1$

Step 5

Solve the equation for $y$ .

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

$y - 6 = \pm \sqrt{1}$

Any root of $1$ is $1$ .

$y - 6 = \pm 1$

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.

$y - 6 = 1$

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

Add $6$ to both sides of the equation.

$y = 1 + 6$

Add $1$ and $6$ .

$y = 7$

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

$y - 6 = - 1$

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

Add $6$ to both sides of the equation.

$y = - 1 + 6$

Add $- 1$ and $6$ .

$y = 5$

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

$y = 7, 5$