Chapter 5 - Polynomials and Factoring - 5.7 Solving Quadratic Equations by Factoring - 5.7 Exercise Set - Page 351: 9

$x=\left\{ -6,\dfrac{3}{2} \right\}$

Equating each factor to zero (Zero Product Principle), then the solutions to the given equation, $ (2t-3)(t+6)=0 ,$ is \begin{array}{l} 2t-3=0 \\\\ 2t=3 \\\\ t=\dfrac{3}{2} ,\\\\\text{OR}\\\\ t+6=0 \\\\ t=-6 .\end{array} Hence, $ x=\left\{ -6,\dfrac{3}{2} \right\} .$