Question

how to solve k^2+3k=108

Answer 1

k^2+3k-108=0
k^2+12k-9k-108=0
k(k+12)-9(k+12)=0
(k-9)(k+12)=0
k=9
and k=-12

Answer 2

k^2 +3k - 108 = 0
(k+12)(k-9)=0
k=9 and k=-12

Answer 3

try factoring as
\[(x-9)(x+12)=0\]

Answer 4

When the coefficient of x^2 is 1, then you can look for 2 numbers that add up to the x-coefficient of x and multiply together to give the constant.

Often works with "made up" problems.

Answer 5

hey estudier r u on facebook

Answer 6

Yes, just so I can log in other places without any headaches..

Answer 7

lolx
hey can u give ur id so that i can add u as a friend

Answer 8

Like I said, I don't use the account, so it's a bit pointless...sorry :-)

Answer 9

oh................np