How do you solve #3t^2 – 4t = –30# by completing the square?

1 Answer
Jul 15, 2017

Answer:

#t=(2+-i sqrt(86))/3#

Explanation:

Given -

#3t^2-4t=-30#

Divide all the terms by 3

#(3t^2)/3-(4t)/3=(-30)/3#

#t^2-4/3t=-10#

divide #-4/3# by 2. square it and add it on both sides

It is #-4/3-:2; -4/3xx1/2=-4/6#

Square the value. It is #(-4/6)=16/36=4/9#

Add #4/9 # to both sides

#t^2-4/3t+4/9=-10+4/9=(-90+4)/9=(-86)/9#

#(t-2/3)^2=-86/9#

#t-2/3=+-sqrt((-86)/9)#

#t-2/3=sqrt(-86)/3#

#t=2/3+-(i sqrt(86))/3#

#t=(2+-i sqrt(86))/3#