randRangeNonZero( -10, 10 ) randRange( -5, 5 ) * 2 + ( X1 % 2 ) ( X1 + X2 ) * -1 X1 * X2 new Polynomial( 0, 2, [C, B, 1], "x" ) POLY.text() $._("or")

Complete the square to solve for `x`.

`POLY_TEXT = 0`

B/2 C * -1 + pow( B / 2, 2 ) X1 X2
B/2 C * -1 + pow( B / 2, 2 ) X2 X1
Completed Square:
`(x + {}` `)^2 = {}`
Solution:
`x = {}``\quad\text{OR}\quad x = {}`
X1

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 + Bx = 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 `B`, half of it would be `B / 2`, and squaring it gives us `\color{blue}{pow( B / 2, 2 )}`.

`x^2 + Bx \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 `B`, half of it is `B / 2`, and squaring it gives us `\color{blue}{pow( B / 2, 2 )}`, our constant term.

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

Isolate `x` to find the solution(s).

`x = -B / 2\pmsqrt( 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 )`