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

Nov 16, 2016

$y = 6$

#### Explanation:

your equation$\setminus \Rightarrow \sqrt[3]{6 y + 2} - 2 = 0$

• add 2 to both sides
$\sqrt[3]{6 y + 2} = 2$
• cube both sides to get rid of cube-root
$6 y + 2 = {2}^{3}$
• simplify the cube
$6 y + 2 = 8$
• subtract 2 from both sides
$6 y = 8 - 2$
• simplify
$6 y = 6$
• divide both sides by 6 to isolate $y$ on left side
$y = 6$