College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 1, Equations and Graphs - Section 1.4 - Solving Quadratic Equations - 1.4 Exercises - Page 123: 13


$z=-\displaystyle \frac{5}{6}$ or $z=\displaystyle \frac{9}{2}$

Work Step by Step

We solve by factoring: $12z^{2}-44z=45$ $12z^{2}-44z-45=0$ $(6z+5)(2z-9)=0$ $6z+5=0$ or $2z-9=0$ So we have: $z=-\displaystyle \frac{5}{6}$ or $z=\displaystyle \frac{9}{2}$
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