How do you solve using the completing the square method `r^2 + 14r = -13`?

`r^2 + 14r = -13`

`implies r^2 + 14r + 13 = 0`

We could have a go at factoring that, but completing the square is the next step down the list and is what you are asking for.

It is the Quadratic Equation without having to remember a formula!

So:

`r^2 + 14r + 13`

`= r^2 + (color(red)(2 xx) 7)r + 13`

[Remember that: `(a + b)^2 = a^2 + color(red)(2) ab + b^2`]

`= (r + 7)^2 - 7^2 + 13`

`= (r + 7)^2 - 36`

Now:

`(r + 7)^2 - 36 color(red)(= 0)`

`implies (r + 7)^2 = 36`

`implies r + 7 = sqrt36`

`implies r =- 7 pm 6 = -1, -13`

We can check that:

`(r+1)(r+13) = r^2 + 13r + r + 13 color(blue)(= r^2 + 14r + 13)`

First, write it in general form:

`r^2+14r+13=0`

Add bracket around the terms that contain `r`, with the constant term outside the bracket (and factor out the `a` coefficient if there is one):

`(r^2+14r)+13=0`

Now, take half of the value of `b` (the coefficient of the term in `r`) and square it. Add this value and subtract it as well, within the bracket.

`(r^2+14r+49 - 49)+13=0`

Move the subtracted `(b/2)^2` term outside the bracket. (This can require some care if there was an `a` value (a coefficient of the `r^2` term greater than 1.)

`(r^2+14r+49) - 49+13=0`

By design, the part on the brackets is a perfect square:

`(r+7)^2 -36 = 0`

It is now in standard form, so you are done.

The vertex is at (-7, -36).

`r^2+14r=-13`

`"Let us rearrange the equation."`

`r^2+14r+13=0`

`"let us add+36 -36"`

`r^2+14rcolor(red)(+36-36)+13=0`

`r^2+14r+36+13color(red)(-36)=0`

`color(green)(r^2+14r+49)-color(red)(36)=0`

`color(green)(r^2+14r+49)=(r+7)^2`

`(r+7)^2-36=0`

`(r+7)^2-6^2=0`

`"let us remember " a^2-b^2=(a-b)(a+b)`

`(r+7-6)(r+7+6)=0`

`(r+1)(r+13)=0`

`"Either (r+1) or (r+13) is equal to zero."`

`r+1=0 rArr r=-1`

`r+13=0rArr r=-13`