How do you solve #(6x+2)^2+4=28#?

1 Answer
Jul 23, 2017

Answer:

#x=-1/3+-1/3sqrt6#

Explanation:

#"isolate " (6x+2)^2#

#"subtract 4 from both sides"#

#rArr(6x+2)^2=24#

#color(blue)"take the square root of both sides"#

#sqrt((6x+2)^2)=+-sqrt24larr" note plus or minus"#

#rArr6x+2=+-2sqrt6larr" simplifying radical"#

#"subtract 2 from both sides"#

#rArr6x=+-2sqrt6-2#

#"divide both sides by 6"#

#rArrx=(-2+-2sqrt6)/6#

#color(white)(rArrx)=-1/3+-1/3sqrt6larrcolor(red)" exact values"#

#x~~ 0.483" to 3 dec. places"#

#"or " x~~-1.15" to 2 dec. places"#