# How do you find all real number solutions to #root3(6y+2)-2=0#?

##### 1 Answer

Nov 16, 2016

#### Answer:

#### Explanation:

your equation

- add 2 to both sides

#root3(6y+2)=2# - cube both sides to get rid of cube-root

#6y+2=2^3# - simplify the cube

#6y+2=8# - subtract 2 from both sides

#6y=8-2# - simplify

#6y=6# - divide both sides by 6 to isolate
#y# on left side

#y=6#