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 - Page 453: 36


{$\frac{-3-\sqrt {13}}{4},\frac{-3+\sqrt {13}}{4}$}

Work Step by Step

Step 1: Comparing $4n^{2}+6n-1=0$ to the standard form of a quadratic equation, $an^{2}+bn+c=0$, we find: $a=4$, $b=6$ and $c=-1$ Step 2: The quadratic formula is: $n=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$ Step 3: Substituting the values of a,b and c in the formula: $n=\frac{-(6) \pm \sqrt {(6)^{2}-4(4)(-1)}}{2(4)}$ Step 4: $n=\frac{-6 \pm \sqrt {36+16}}{8}$ Step 5: $n=\frac{-6 \pm \sqrt {52}}{8}$ Step 6: $n=\frac{-6 \pm \sqrt {4\times13}}{8}$ Step 7: $n=\frac{-6 \pm 2\sqrt {13}}{8}$ Step 8: $n=\frac{2(-3 \pm \sqrt {13})}{8}$ Step 9: $n=\frac{(-3 \pm \sqrt {13})}{4}$ Step 10: $n=\frac{-3-\sqrt {13}}{4}$ or $n=\frac{-3+\sqrt {13}}{4}$ Step 11: Therefore, the solution set is {$\frac{-3-\sqrt {13}}{4},\frac{-3+\sqrt {13}}{4}$}.
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.