How do you solve #x(x-6)=72#?

Question

1531 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-10T17:52:19

#x=-6,12#

The first step is to distribute #x(x-6)#:

#x^2-6x#

Now you have:

#x^2-6x=72#

Subtract 72 from both sides:

#x^2-6x-72=cancel72-cancel72#

#x^2-6x-72=0#

Factor it:

#(x-12)(x+6)=0#

Set each of them equal to zero and solve:

#x-12=0#

#x+6=0#

The answer is:

#x=-6,12#