How do you solve `x^2+3x+21=22` by completing the square?

Steps:

Subtract 21 from both sides

`x^2+3x + ? = 1`

In order to find ?, you must divide `3` by `2` or `(b/2)`and square it and add it to both sides:

`? = (3/2)^2 =9/4`

`x^2+3x + 9/4 = 1 +9/4`

Now we can factor the left side and simplify the right.

`(x+3/2)^2 = 4/4+9/4`

`(x+3/2)^2 = 13/4`

Finally, solve for `x`

`x+3/2=0 -> x=-3/2`

`x^2-3/14=0 -> x^2 = 13/14 -> x = +- sqrt(13/4) -> x = +- sqrt(13)/2`

Squares of form `(x+a)^2` are expanded like this:

`x^2 + 2ax + a^2`

Since our equation starts with `x^2 + 3x...`, we know that `3 = 2a`, so `3/2 = a`. That means that to complete the square, we need the last term to be `a^2 = (3/2)^2 = 9/4`.

To do this, we can subtract `18.75` from both sides and simplify.

`color(white)"X"x^2 + 3x + 21 = 22`

`color(white)"X" x^2 + 3x + 9/4 = 13/4`

`color(white)"XX.." (x + 1.5)^2 = 13/4`

Now we can take the square root of both sides. Remember that the square root can be either positive or negative, since when squared, the negative sign disappears.

`color(white)"" x + 3/2 = +- sqrt13/2`

`color(white)"XXX." x = (-3 +- sqrt13)/2`

Final Answer