## Thomas' Calculus 13th Edition

Solution set = $\displaystyle \{-\frac{1}{2},-\frac{9}{2}\}$
Use properties from the table 'Absolute Values and Intervals", Here: property 5: $|x|=a \Leftrightarrow x=\pm a,\ \quad$(for any positive number a) Thus: $2t+5=\pm 4\qquad$... add $-5$ $2t=-5\pm 4\qquad$... divide with 2, $t=\displaystyle \frac{-5\pm 4}{2}$ $\displaystyle \frac{-5+4}{2}=-\frac{1}{2}$ $\displaystyle \frac{-5-4}{2}=-\frac{9}{2}$ Solution set = $\displaystyle \{-\frac{1}{2},-\frac{9}{2}\}$