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

Jan 22, 2017

The answer is $S = \emptyset$

#### Explanation:

Let's rewrite the inequality

${t}^{2} < 3 \left(2 t - 3\right)$

${t}^{2} - 3 \left(2 t - 3\right) < 0$

${t}^{2} - 6 t + 9 < 0$

Let's factorise

$\left(t - 3\right) \left(t - 3\right) < 0$

${\left(t - 3\right)}^{2} < 0$

This is impossible since ${\left(t - 3\right)}^{2} \ge 0 , \forall x \in \mathbb{R}$