**Completed Square:**

`(x + {}`

`)^2 = {}`

**Solution:**

`x = {}`

`\quad\text{``OR`}\quad x = {}

Complete the square to solve for `x`

.

`POLY_TEXT` = 0

`(x + {}`

`)^2 = {}`

`x = {}`

`\quad\text{``OR`}\quad x = {}

**Completed Square:**

`(x + {}`

`-X1` `)^2 = {}`

0

**Solution:**

`x = \quad`

`X1`

Begin by moving the constant term to the right side of the equation.

`x^2 + `

`B`x = `C * -1`

We complete the square by taking half of the coefficient of our `x`

term, squaring it, and adding it to both sides of the equation. Since the coefficient of our `x`

term is

, half of it would be `B`

, and squaring it gives us `B / 2``\color{blue}{`

.`pow( B / 2, 2 )`}

`x^2 + `

`B`x \color{blue}{ + `pow( B / 2, 2 )`} = `C * -1` \color{blue}{ + `pow( B / 2, 2 )`}

We can now rewrite the left side of the equation as a squared term.

`( x + `

`B / 2` )^2 = `C * -1 + pow( B / 2, 2 )`

The left side of the equation is already a perfect square trinomial. The coefficient of our `x`

term is

, half of it is `B`

, and squaring it gives us `B / 2``\color{blue}{`

, our constant term.`pow( B / 2, 2 )`}

Thus, we can rewrite the left side of the equation as a squared term.

`( x + `

`B / 2` )^2 = `C * -1 + pow( B / 2, 2 )`

Take the square root of both sides.

`x + `

`B / 2` = \pm`sqrt( C * -1 + pow( B / 2, 2 ) )`

Isolate `x`

to find the solution(s).

`x = `

`-B / 2`\pm`sqrt( C * -1 + pow( B / 2, 2 ) )`

So the solutions are: `x = `

`-B / 2 + sqrt( C * -1 + pow( B / 2, 2 ) )` \text{ `OR` } x = `-B / 2 - sqrt( C * -1 + pow( B / 2, 2 ) )`

The solution is: `x = `

`-B / 2 + sqrt( C * -1 + pow( B / 2, 2 ) )`

We already found the completed square: `( x + `

`B / 2` )^2 = `C * -1 + pow( B / 2, 2 )`