Question

How do you solve `r^2+3r-3=0` by completing the square?

Answer 1

I found:
`r_1=(-3+sqrt(21))/2`
`r_2=(-3-sqrt(21))/2`

Explanation:

Write it as:
`r^2+3r=3`
add and subtract `9/4` (i.e., a number that comes from `3/2` so that squared gives you `9/4` and multiplied by `2` gives you `3` which takes care of `3r`) to get:
`r^2+3rcolor(red)(+9/4-9/4)=3` rearranging:
`r^2+3r+9/4=3+9/4`
`(r+3/2)^2=(12+9)/4`
`(r+3/2)^2=21/4` square root both sides:
`r+3/2=+-sqrt(21/4)` you get two solutions:
`r_1=(-3+sqrt(21))/2`
`r_2=(-3-sqrt(21))/2`