How do you solve #t^2<3(2t-3)# using a sign chart?

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Answer 1 · 2017-01-22T11:38:20

The answer is #S=O/#

Let's rewrite the inequality

#t^2<3(2t-3)#

#t^2-3(2t-3)<0#

#t^2-6t+9<0#

Let's factorise

#(t-3)(t-3)<0#

#(t-3)^2<0#

This is impossible since #(t-3)^2>=0, AA x in RR#