Question

How do you solve `k^ { 2} + 23k + 12= - 120`?

Answer 1

`k=-11`
`or`
`k=-12`

Explanation:

`k^2+23k+12=-120`

`k^2+23k+12+120=0`

`k^2+23k+132=0`

`(k+11)(k+12)=0`

`k+11=0` ==> `k=-11`
`or`
`k+12=0` ==> `k=-12`