Solve by Completing the Square y^2-4y-5=0

$y^{2} - 4 y - 5 = 0$

Step 1

Add $5$ to both sides of the equation.

$y^{2} - 4 y = 5$

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.

$y^{2} - 4 y + {(- 2)}^{2} = 5 + {(- 2)}^{2}$

Step 4

Simplify the equation.

Simplify the left side.

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

$y^{2} - 4 y + 4 = 5 + {(- 2)}^{2}$

Simplify the right side.

Simplify $5 + {(- 2)}^{2}$ .

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

$y^{2} - 4 y + 4 = 5 + 4$

Add $5$ and $4$ .

$y^{2} - 4 y + 4 = 9$

Step 5

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

${(y - 2)}^{2} = 9$

Step 6

Solve the equation for $y$ .

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

$y - 2 = \pm \sqrt{9}$

Simplify $\pm \sqrt{9}$ .

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

$y - 2 = \pm \sqrt{3^{2}}$

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

$y - 2 = \pm 3$

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.

$y - 2 = 3$

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

Add $2$ to both sides of the equation.

$y = 3 + 2$

Add $3$ and $2$ .

$y = 5$

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

$y - 2 = - 3$

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

Add $2$ to both sides of the equation.

$y = - 3 + 2$

Add $- 3$ and $2$ .

$y = - 1$

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

$y = 5, - 1$

Solve by Completing the Square y^2-4y-5=0

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