Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 10 - Quadratic Equations - 10.3 - Quadratic Formula - Problem Set 10.3: 27

Answer

{$-\frac{5}{4},-\frac{1}{3}$}

Work Step by Step

Step 1: Comparing $12x^{2}+19x+5=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we obtain: $a=12$, $b=19$, and $c=5$ Step 2: The quadratic formula is: $x=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$ Step 3: Substituting the values of a,b, and c in the formula, we obtain: $x=\frac{-(19) \pm \sqrt {(19)^{2}-4(12)(5)}}{2(12)}$ Step 4: $x=\frac{-19 \pm \sqrt {361-240}}{24}$ Step 5: $x=\frac{-19 \pm \sqrt {121}}{24}$ Step 6: $x=\frac{-19 \pm 11}{24}$ Step 7: $x=\frac{-19+11}{24}$ or $x=\frac{-19-11}{24}$ Step 8: $x=\frac{-8}{24}$ or $x=\frac{-30}{24}$ Step 9: $x=-\frac{1}{3}$ or $x=-\frac{5}{4}$ Step 10: Therefore, the solution set is {$-\frac{5}{4},-\frac{1}{3}$}.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.