How do you solve #w(6w+1)=2#?

Question

1590 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-15T02:06:29

First distribute the parentheses.

#6w^2 + w = 2#

#6w^2 + w - 2 = 0#

#6w^2 + 4w - 3w - 2 = 0#

#2w(3w + 2) - 1(3w + 2) = 0#

#(2w - 1)(3w + 2) = 0#

#w = 1/2 and -2/3#

Hopefully this helps!